130) main()
{
while
(strcmp(“some”,”some\0”))
printf(“Strings
are not equal\n”);
}
Answer:
No
output
Explanation:
Ending the
string constant with \0 explicitly makes no difference. So “some” and “some\0”
are equivalent. So, strcmp returns 0 (false) hence breaking out of the while
loop.
131) main()
{
char
str1[] = {‘s’,’o’,’m’,’e’};
char
str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while
(strcmp(str1,str2))
printf(“Strings
are not equal\n”);
}
Answer:
“Strings
are not equal”
“Strings
are not equal”
….
Explanation:
If a string
constant is initialized explicitly with characters, ‘\0’ is not appended
automatically to the string. Since str1 doesn’t have null termination, it
treats whatever the values that are in the following positions as part of the
string until it randomly reaches a ‘\0’. So str1 and str2 are not the same,
hence the result.
132) main()
{
int
i = 3;
for
(;i++=0;) printf(“%d”,i);
}
Answer:
Compiler
Error: Lvalue required.
Explanation:
As we know that
increment operators return rvalues and
hence it cannot appear on the left hand side of an assignment operation.
133) void main()
{
int
*mptr, *cptr;
mptr
= (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int
*cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value
0
Explanation:
The memory
space allocated by malloc is uninitialized, whereas calloc returns the
allocated memory space initialized to zeros.
134) void main()
{
static
int i;
while(i<=10)
(i>2)?i++:i--;
printf(“%d”, i);
}
Answer: 32767
Explanation:
Since i is
static it is initialized to 0. Inside the while loop the conditional operator
evaluates to false, executing i--. This continues till the integer value
rotates to positive value (32767). The while condition becomes false and hence,
comes out of the while loop, printing the i value.
135) main()
{
int
i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d
%d",i,j);
}
Answer:
10
10
Explanation:
The
Ternary operator ( ? : ) is equivalent for if-then-else statement. So the
question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j
= j;
136) 1. const char *a;
2. char* const
a;
3. char const
*a;
-Differentiate
the above declarations.
Answer:
1. 'const'
applies to char * rather than 'a' ( pointer to a constant char )
*a='F' : illegal
a="Hi" : legal
2.
'const' applies to 'a' rather than to
the value of a (constant pointer to char )
*a='F' : legal
a="Hi" : illegal
3.
Same as 1.
137) main()
{
int
i=5,j=10;
i=i&=j&&10;
printf("%d
%d",i,j);
}
Answer:1 10
Explanation:
The expression
can be written as i=(i&=(j&&10)); The inner expression
(j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence
the result.
138) main()
{
int
i=4,j=7;
j = j || i++ &&
printf("YOU CAN");
printf("%d
%d", i, j);
}
Answer:
4
1
Explanation:
The boolean expression needs to be evaluated
only till the truth value of the expression is not known. j is not equal to
zero itself means that the expression’s truth value is 1. Because it is
followed by || and true || (anything)
=> true where (anything) will not be evaluated. So the remaining
expression is not evaluated and so the value of i remains the same.
Similarly when
&& operator is involved in an expression, when any of the operands
become false, the whole expression’s truth value becomes false and hence the
remaining expression will not be evaluated.
false
&& (anything) => false where (anything) will not be evaluated.
139) main()
{
register
int a=2;
printf("Address of a =
%d",&a);
printf("Value
of a = %d",a);
}
Answer:
Compier
Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied
on register variables.
140) main()
{
float
i=1.5;
switch(i)
{
case 1:
printf("1");
case
2: printf("2");
default
: printf("0");
}
}
Answer:
Compiler
Error: switch expression not integral
Explanation:
Switch statements can be applied only to
integral types.
141) main()
{
extern
i;
printf("%d\n",i);
{
int
i=20;
printf("%d\n",i);
}
}
Answer: Linker Error : Unresolved
external symbol i
Explanation:
The identifier
i is available in the inner block and so using extern has no use in resolving
it.
142) main()
{
int
a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d %d
%d",a,*f1,*f2);
}
Answer:
16
16 16
Explanation:
f1 and f2 both
refer to the same memory location a. So changes through f1 and f2 ultimately
affects only the value of a.
143) main()
{
char
*p="GOOD";
char a[ ]="GOOD";
printf("\n
sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p), sizeof(*p),
strlen(p));
printf("\n sizeof(a) = %d,
strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
sizeof(p)
= 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a) = 5, strlen(a) = 4
Explanation:
sizeof(p)
=> sizeof(char*) => 2
sizeof(*p) => sizeof(char) =>
1
Similarly,
sizeof(a) => size of the
character array => 5
When sizeof
operator is applied to an array it returns the sizeof the array and it is not the same as the sizeof the pointer
variable. Here the sizeof(a) where a is the character array and the size of the
array is 5 because the space necessary for the terminating NULL character
should also be taken into account.
144) #define DIM( array, type)
sizeof(array)/sizeof(type)
main()
{
int arr[10];
printf(“The dimension of the array is %d”,
DIM(arr, int));
}
Answer:
10
Explanation:
The size of integer array of 10 elements is 10 *
sizeof(int). The macro expands to sizeof(arr)/sizeof(int) => 10 *
sizeof(int) / sizeof(int) => 10.
145) int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(“The dimension of the array is %d”,
DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as
arguments and only the pointers can be passed. So the argument is
equivalent to int * array (this is one of the very few places where [] and *
usage are equivalent). The return statement becomes, sizeof(int *)/ sizeof(int)
that happens to be equal in this case.
146) main()
{
static int
a[3][3]={1,2,3,4,5,6,7,8,9};
int i,j;
static *p[]={a,a+1,a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),
*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
}
}
Answer:
1 1
1 1
2 4
2 4
3 7
3 7
4 2
4 2
5 5
5 5
6 8
6 8
7 3
7 3
8 6
8 6
9 9
9 9
Explanation:
*(*(p+i)+j)
is equivalent to p[i][j].
147) main()
{
void
swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int
*a, int *b)
{
*a ^= *b,
*b ^= *a, *a ^= *b;
}
Answer:
x=10
y=8
Explanation:
Using ^ like
this is a way to swap two variables without using a temporary variable and that
too in a single statement.
Inside main(),
void swap(); means that swap is a function that may take any number of
arguments (not no arguments) and returns nothing. So this doesn’t issue a
compiler error by the call swap(&x,&y); that has two arguments.
This convention
is historically due to pre-ANSI style (referred to as Kernighan and Ritchie
style) style of function declaration. In that style, the swap function will be
defined as follows,
void swap()
int *a, int *b
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
where the
arguments follow the (). So naturally the declaration for swap will look like,
void swap() which means the swap can take any number of arguments.
148) main()
{
int
i = 257;
int *iPtr = &i;
printf("%d
%d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
1
1
Explanation:
The integer
value 257 is stored in the memory as, 00000001 00000001, so the individual
bytes are taken by casting it to char * and get printed.
149) main()
{
int
i = 258;
int *iPtr = &i;
printf("%d
%d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
2
1
Explanation:
The integer
value 257 can be represented in binary as, 00000001 00000001. Remember that the
INTEL machines are ‘small-endian’ machines. Small-endian
means that the lower order bytes are stored in the higher memory addresses and
the higher order bytes are stored in lower addresses. The integer value 258
is stored in memory as: 00000001 00000010.
150) main()
{
int
i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}
Answer:
556
Explanation:
The integer
value 300 in binary notation is:
00000001 00101100. It is stored in
memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr =
2 makes the memory representation as: 00101100 00000010. So the integer
corresponding to it is 00000010 00101100 => 556.
151) #include <stdio.h>
main()
{
char
* str = "hello";
char
* ptr = str;
char
least = 127;
while
(*ptr++)
least = (*ptr<least )
?*ptr :least;
printf("%d",least);
}
Answer:
0
Explanation:
After ‘ptr’
reaches the end of the string the value pointed by ‘str’ is ‘\0’. So the value
of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.
152) Declare an array of N pointers to functions
returning pointers to functions returning pointers to characters?
Answer:
(char*(*)(
)) (*ptr[N])( );
153) main()
{ struct student
{
char name[30];
struct date dob;
}stud;
struct
date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno,
&student.dob.day, &student.dob.month, &student.dob.year);
}
Answer:
Compiler
Error: Undefined structure date
Explanation:
Inside the
struct definition of ‘student’ the member of type struct date is given. The
compiler doesn’t have the definition of date structure (forward reference is not allowed in C in this case)
so it issues an error.
154) main()
{
struct
date;
struct
student
{
char
name[30];
struct
date dob;
}stud;
struct
date
{
int day,month,year;
};
scanf("%s%d%d%d",
stud.rollno, &student.dob.day, &student.dob.month,
&student.dob.year);
}
Answer:
Compiler
Error: Undefined structure date
Explanation:
Only
declaration of struct date is available inside the structure definition of
‘student’ but to have a variable of type struct date the definition of the
structure is required.
155) There were 10 records stored in
“somefile.dat” but the following program printed 11 names. What went wrong?
void main()
{
struct
student
{
char
name[30], rollno[6];
}stud;
FILE
*fp = fopen(“somefile.dat”,”r”);
while(!feof(fp))
{
fread(&stud,
sizeof(stud), 1 , fp);
puts(stud.name);
}
}
Explanation:
fread reads 10
records and prints the names successfully. It will return EOF only when fread
tries to read another record and fails reading EOF (and returning EOF). So it
prints the last record again. After this only the condition feof(fp) becomes
false, hence comes out of the while loop.
156) Is there any difference between the two declarations,
1.
int foo(int *arr[]) and
2.
int foo(int *arr[2])
Answer:
No
Explanation:
Functions can
only pass pointers and not arrays. The numbers that are allowed inside the []
is just for more readability. So there is no difference between the two
declarations.
157) What is the subtle error in the following code segment?
void fun(int n,
int arr[])
{
int
*p=0;
int
i=0;
while(i++<n)
p = &arr[i];
*p = 0;
}
Answer & Explanation:
If the body of
the loop never executes p is assigned no address. So p remains NULL where *p =0
may result in problem (may rise to runtime error “NULL pointer assignment” and
terminate the program).
158) What is wrong with the following code?
int *foo()
{
int
*s = malloc(sizeof(int)100);
assert(s
!= NULL);
return
s;
}
Answer & Explanation:
assert macro
should be used for debugging and finding out bugs. The check s != NULL is for
error/exception handling and for that assert shouldn’t be used. A plain if and
the corresponding remedy statement has to be given.
159) What is the hidden bug with the following statement?
assert(val++
!= 0);
Answer & Explanation:
Assert macro is
used for debugging and removed in release version. In assert, the experssion
involves side-effects. So the behavior of the code becomes different in case of
debug version and the release version thus leading to a subtle bug.
Rule to Remember:
Don’t use expressions that have side-effects
in assert statements.
160) void main()
{
int *i =
0x400; // i points to the address 400
*i = 0; // set the value of memory
location pointed by i;
}
Answer:
Undefined
behavior
Explanation:
The second
statement results in undefined behavior because it points to some location
whose value may not be available for modification. This
type of pointer in which the non-availability of the implementation of the
referenced location is known as 'incomplete type'.
161) #define assert(cond) if(!(cond)) \
(fprintf(stderr, "assertion failed: %s,
file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort())
void main()
{
int i = 10;
if(i==0)
assert(i < 100);
else
printf("This statement becomes else
for if in assert macro");
}
Answer:
No
output
Explanation:
The else part in
which the printf is there becomes the else for if in the assert macro. Hence
nothing is printed.
The solution is
to use conditional operator instead of if statement,
#define
assert(cond) ((cond)?(0): (fprintf (stderr, "assertion failed: \ %s, file
%s, line %d \n",#cond, __FILE__,__LINE__), abort()))
Note:
However this
problem of “matching with nearest else” cannot be solved by the usual method of
placing the if statement inside a block like this,
#define
assert(cond) { \
if(!(cond))
\
(fprintf(stderr, "assertion failed: %s,
file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort()) \
}
162) Is the following code legal?
struct a
{
int x;
struct a b;
}
Answer: No
Explanation:
Is
it not legal for a structure to contain a member that is of the same
type as in this
case. Because this will cause the structure declaration to be recursive without
end.
163) Is the following code legal?
struct a
{
int
x;
struct a *b;
}
Answer:
Yes.
Explanation:
*b is a pointer
to type struct a and so is legal. The compiler knows, the size of the pointer
to a structure even before the size of the structure
is
determined(as you know the pointer to any type is of same size). This type of
structures is known as ‘self-referencing’ structure.
164) Is the following code legal?
typedef struct a
{
int
x;
aType *b;
}aType
Answer:
No
Explanation:
The typename
aType is not known at the point of declaring the structure (forward references
are not made for typedefs).
165) Is the following code legal?
typedef struct a
aType;
struct a
{
int
x;
aType
*b;
};
Answer:
Yes
Explanation:
The typename
aType is known at the point of declaring the structure, because it is already
typedefined.
166) Is the following code legal?
void main()
{
typedef
struct a aType;
aType
someVariable;
struct
a
{
int
x;
aType *b;
};
}
Answer:
No
Explanation:
When
the declaration,
typedef
struct a aType;
is encountered
body of struct a is not known. This is known as ‘incomplete types’.
167) void main()
{
printf(“sizeof
(void *) = %d \n“, sizeof( void *));
printf(“sizeof
(int *) = %d \n”, sizeof(int *));
printf(“sizeof
(double *) = %d \n”, sizeof(double *));
printf(“sizeof(struct
unknown *) = %d \n”, sizeof(struct unknown *));
}
Answer :
sizeof
(void *) = 2
sizeof
(int *) = 2
sizeof
(double *) = 2
sizeof(struct
unknown *) = 2
Explanation: The pointer to any type is
of same size.
168) char inputString[100] = {0};
To get string
input from the keyboard which one of the following is better?
1) gets(inputString)
2) fgets(inputString,
sizeof(inputString), fp)
Answer & Explanation:
The second one
is better because gets(inputString) doesn't know the size of the string passed
and so, if a very big input (here, more than 100 chars) the charactes will be
written past the input string. When fgets is used with stdin performs the same
operation as gets but is safe.
169) Which version do you prefer of the following two,
1)
printf(“%s”,str); // or the more curt
one
2)
printf(str);
Answer & Explanation:
Prefer the
first one. If the str contains any
format characters like %d then it will result in a subtle bug.
170) void main()
{
int
i=10, j=2;
int
*ip= &i, *jp = &j;
int
k = *ip/*jp;
printf(“%d”,k);
}
Answer:
Compiler
Error: “Unexpected end of file in comment started in line 5”.
Explanation:
The programmer
intended to divide two integers, but by the “maximum munch” rule, the compiler
treats the operator sequence / and * as /* which happens to be the starting of
comment. To force what is intended by the programmer,
int k = *ip/ *jp;
// give space explicity separating / and *
//or
int k = *ip/(*jp);
// put braces to force the intention
will solve the problem.
171) void main()
{
char ch;
for(ch=0;ch<=127;ch++)
printf(“%c %d \n“, ch, ch);
}
Answer:
Implementaion dependent
Explanation:
The char type
may be signed or unsigned by default. If it is signed then ch++ is executed
after ch reaches 127 and rotates back to -128. Thus ch is always smaller than
127.
172) Is
this code legal?
int *ptr;
ptr = (int *)
0x400;
Answer:
Yes
Explanation:
The
pointer ptr will point at the integer in the memory location 0x400.
173) main()
{ char a[4]="HELLO";
printf("%s",a);
}
Answer:
Compiler
error: Too many initializers
Explanation:
The array a is
of size 4 but the string constant requires 6 bytes to get stored.
174) main()
{ char a[4]="HELL";
printf("%s",a);
}
Answer:
HELL%@!~@!@???@~~!
Explanation:
The character
array has the memory just enough to hold the string “HELL” and doesnt have
enough space to store the terminating null character. So it prints the HELL
correctly and continues to print garbage values till it accidentally comes across a NULL character.
175) main()
{
int
a=10,*j;
void *k;
j=k=&a;
j++;
k++;
printf("\n
%u %u ",j,k);
}
Answer:
Compiler
error: Cannot increment a void pointer
Explanation:
Void pointers
are generic pointers and they can be used only when the type is not known and
as an intermediate address storage type. No pointer arithmetic can be done on
it and you cannot apply indirection operator (*) on void pointers.
176) main()
{
extern
int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
int
i;
177) Printf can be implemented by using __________ list.
Answer:
Variable
length argument lists
178) char
*someFun()
{
char *temp = “string constant";
return temp;
}
int main()
{
puts(someFun());
}
Answer:
string constant
Explanation:
The program suffers no problem and
gives the output correctly because the character constants are stored in
code/data area and not allocated in stack, so this doesn’t lead to dangling
pointers.
179) char *someFun1()
{
char temp[ ] = “string";
return temp;
}
char *someFun2()
{
char temp[ ] = {‘s’,
‘t’,’r’,’i’,’n’,’g’};
return temp;
}
int main()
{
puts(someFun1());
puts(someFun2());
}
Answer:
Garbage
values.
Explanation:
Both
the functions suffer from the problem of dangling pointers. In someFun1() temp
is a character array and so the space for it is allocated in heap and is
initialized with character string “string”. This is created dynamically as the
function is called, so is also deleted dynamically on exiting the function so
the string data is not available in the calling function main() leading to
print some garbage values. The function someFun2() also suffers from the same problem
but the problem can be easily identified in this case.
C++ Aptitude
and OOPS
C++
Aptitude and OOPS
Note : All the programs are tested under Turbo C++ 3.0, 4.5 and Microsoft
VC++ 6.0 compilers.
It is assumed
that,
Ø Programs
run under Windows environment,
Ø The
underlying machine is an x86 based system,
Ø Program
is compiled using Turbo C/C++ compiler.
The program
output may depend on the information based on this assumptions (for example
sizeof(int) == 2 may be assumed).
1) class Sample
{
public:
int *ptr;
Sample(int i)
{
ptr = new int(i);
}
~Sample()
{
delete ptr;
}
void PrintVal()
{
cout << "The value is "
<< *ptr;
}
};
void
SomeFunc(Sample x)
{
cout <<
"Say i am in someFunc " << endl;
}
int main()
{
Sample s1= 10;
SomeFunc(s1);
s1.PrintVal();
}
Answer:
Say i am in
someFunc
Null pointer
assignment(Run-time error)
Explanation:
As the object is
passed by value to SomeFunc the
destructor of the object is called when the control returns from the function.
So when PrintVal is called it meets up with ptr
that has been freed.The solution is to pass the Sample object by
reference to SomeFunc:
void
SomeFunc(Sample &x)
{
cout <<
"Say i am in someFunc " << endl;
}
because when we pass objects by refernece that object is not destroyed.
while returning from the function.
2)
Which is the parameter that is added to every
non-static member function when it is called?
Answer:
‘this’
pointer
3) class base
{
public:
int bval;
base(){ bval=0;}
};
class deri:public base
{
public:
int dval;
deri(){ dval=1;}
};
void SomeFunc(base *arr,int size)
{
for(int i=0; i<size;
i++,arr++)
cout<<arr->bval;
cout<<endl;
}
int main()
{
base BaseArr[5];
SomeFunc(BaseArr,5);
deri DeriArr[5];
SomeFunc(DeriArr,5);
}
Answer:
00000
01010
Explanation:
The function
SomeFunc expects two arguments.The first one is a pointer to an array of base
class objects and the second one is the sizeof the array.The first call of
someFunc calls it with an array of bae objects, so it works correctly and
prints the bval of all the objects. When Somefunc is called the second time the
argument passed is the pointeer to an array of derived class objects and not
the array of base class objects. But that is what the function expects to be
sent. So the derived class pointer is promoted to base class pointer and the
address is sent to the function. SomeFunc() knows nothing about this and just
treats the pointer as an array of base class objects. So when arr++ is met, the
size of base class object is taken into consideration and is incremented by
sizeof(int) bytes for bval (the deri class objects have bval and dval as members
and so is of size >= sizeof(int)+sizeof(int) ).
4) class base
{
public:
void
baseFun(){ cout<<"from base"<<endl;}
};
class deri:public base
{
public:
void
baseFun(){ cout<< "from derived"<<endl;}
};
void SomeFunc(base *baseObj)
{
baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from
base
from
base
Explanation:
As
we have seen in the previous case, SomeFunc expects a pointer to a base class.
Since a pointer to a derived class object is passed, it treats the argument
only as a base class pointer and the corresponding base function is called.
5) class base
{
public:
virtual
void baseFun(){ cout<<"from base"<<endl;}
};
class deri:public base
{
public:
void
baseFun(){ cout<< "from derived"<<endl;}
};
void SomeFunc(base *baseObj)
{
baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from
base
from
derived
Explanation:
Remember
that baseFunc is a virtual function. That means that it supports run-time
polymorphism. So the function corresponding to the derived class object is
called.
void
main()
{
int a, *pa, &ra;
pa = &a;
ra = a;
cout <<"a="<<a
<<"*pa="<<*pa <<"ra"<<ra ;
}
/*
Answer
:
Compiler Error: 'ra',reference must
be initialized
Explanation
:
Pointers are different from
references. One of the main
differences
is that the pointers can be both initialized and assigned,
whereas
references can only be initialized. So this code issues an error.
*/
const
int size = 5;
void
print(int *ptr)
{
cout<<ptr[0];
}
void
print(int ptr[size])
{
cout<<ptr[0];
}
void
main()
{
int a[size] = {1,2,3,4,5};
int *b = new int(size);
print(a);
print(b);
}
/*
Answer:
Compiler Error : function 'void
print(int *)' already has a body
Explanation:
Arrays cannot be passed to
functions, only pointers (for arrays, base addresses)
can
be passed. So the arguments int *ptr and int prt[size] have no difference
as
function arguments. In other words, both the functoins have the same signature
and
so
cannot be overloaded.
*/
class
some{
public:
~some()
{
cout<<"some's
destructor"<<endl;
}
};
void
main()
{
some s;
s.~some();
}
/*
Answer:
some's destructor
some's destructor
Explanation:
Destructors can be called
explicitly. Here 's.~some()' explicitly calls the
destructor
of 's'. When main() returns, destructor of s is called again,
hence
the result.
*/
#include
<iostream.h>
class
fig2d
{
int dim1;
int dim2;
public:
fig2d() { dim1=5; dim2=6;}
virtual void
operator<<(ostream & rhs);
};
void
fig2d::operator<<(ostream &rhs)
{
rhs
<<this->dim1<<" "<<this->dim2<<"
";
}
/*class
fig3d : public fig2d
{
int dim3;
public:
fig3d() { dim3=7;}
virtual void
operator<<(ostream &rhs);
};
void
fig3d::operator<<(ostream &rhs)
{
fig2d::operator <<(rhs);
rhs<<this->dim3;
}
*/
void
main()
{
fig2d obj1;
// fig3d obj2;
obj1 << cout;
// obj2 << cout;
}
/*
Answer
:
5 6
Explanation:
In this program, the <<
operator is overloaded with ostream as argument.
This
enables the 'cout' to be present at the right-hand-side. Normally, 'cout'
is
implemented as global function, but it doesn't mean that 'cout' is not possible
to
be overloaded as member function.
Overloading << as virtual member
function becomes handy when the class in which
it
is overloaded is inherited, and this becomes available to be overrided. This is
as opposed
to
global friend functions, where friend's are not inherited.
*/
class
opOverload{
public:
bool operator==(opOverload temp);
};
bool
opOverload::operator==(opOverload temp){
if(*this == temp ){
cout<<"The
both are same objects\n";
return
true;
}
else{
cout<<"The
both are different\n";
return false;
}
}
void
main(){
opOverload a1, a2;
a1= =a2;
}
Answer
:
Runtime Error: Stack Overflow
Explanation
:
Just like normal functions, operator
functions can be called recursively. This program just illustrates that point,
by calling the operator == function recursively, leading to an infinite loop.
class
complex{
double re;
double im;
public:
complex() : re(1),im(0.5) {}
bool operator==(complex &rhs);
operator int(){}
};
bool
complex::operator == (complex &rhs){
if((this->re == rhs.re)
&& (this->im == rhs.im))
return true;
else
return false;
}
int
main(){
complex c1;
cout<< c1;
}
Answer
: Garbage value
Explanation:
The programmer wishes to print the
complex object using output
re-direction
operator,which he has not defined for his lass.But the compiler instead of
giving an error sees the conversion function
and
converts the user defined object to standard object and prints
some
garbage value.
class
complex{
double re;
double im;
public:
complex() : re(0),im(0) {}
complex(double n) { re=n,im=n;};
complex(int m,int n) { re=m,im=n;}
void print() { cout<<re;
cout<<im;}
};
void
main(){
complex c3;
double i=5;
c3 = i;
c3.print();
}
Answer:
5,5
Explanation:
Though no operator= function taking
complex, double is defined, the double on the rhs is converted into a temporary
object using the single argument constructor taking double and assigned to the
lvalue.
void
main()
{
int a, *pa, &ra;
pa = &a;
ra = a;
cout <<"a="<<a
<<"*pa="<<*pa <<"ra"<<ra ;
}
Answer
:
Compiler Error: 'ra',reference must
be initialized
Explanation
:
Pointers are different from
references. One of the main
differences
is that the pointers can be both initialized and assigned,
whereas
references can only be initialized. So this code issues an error.
Try it Yourself
1) Determine the output of the
'C++' Codelet.
class
base
{
public
:
out()
{
cout<<"base
";
}
};
class
deri{
public
: out()
{
cout<<"deri
";
}
};
void
main()
{
deri dp[3];
base
*bp = (base*)dp;
for
(int i=0; i<3;i++)
(bp++)->out();
}
2)
Justify the use of virtual constructors and destructors
in C++.
3)
Each C++ object possesses the 4 member fns,(which can
be declared by the programmer explicitly or by the implementation if they are
not available). What are those 4 functions?
4)
What is wrong
with this class declaration?
class
something
{
char
*str;
public:
something(){
st = new char[10]; }
~something()
{
delete
str;
}
};
5) Inheritance is also known as
-------- relationship. Containership as
________ relationship.
6) When is it necessary to use member-wise
initialization list (also known as
header initialization list) in C++?
7) Which is the only operator in
C++ which can be overloaded but NOT inherited.
8) Is there anything wrong with
this C++ class declaration?
class
temp
{
int value1;
mutable int value2;
public :
void
fun(int val)
const{
((temp*)
this)->value1 = 10;
value2
= 10;
} };
1. What is a modifier?
Answer:
A modifier, also called a modifying
function is a member function that changes the value of at least one data
member. In other words, an operation that modifies the state of an object.
Modifiers are also known as ‘mutators’.
2. What is an accessor?
Answer:
An accessor is a class operation that
does not modify the state of an object. The accessor functions need to be
declared as const operations
3. Differentiate between a
template class and class template.
Answer:
Template class:
A generic
definition or a parameterized class not instantiated until the client provides
the needed information. It’s jargon for plain templates.
Class template:
A class template
specifies how individual classes can be constructed much like the way a class
specifies how individual objects can be constructed. It’s jargon for plain
classes.
4. When does a name clash
occur?
Answer:
A
name clash occurs when a name is defined in more than one place. For example.,
two different class libraries could give two different classes the same name.
If you try to use many class libraries at the same time, there is a fair chance
that you will be unable to compile or link the program because of name clashes.
5. Define namespace.
Answer:
It
is a feature in c++ to minimize name collisions in the global name space. This
namespace keyword assigns a distinct name to a library that allows other
libraries to use the same identifier names without creating any name
collisions. Furthermore, the compiler uses the namespace signature for
differentiating the definitions.
6. What is the use of ‘using’
declaration.
Answer:
A
using declaration makes it possible to use a name from a namespace without the
scope operator.
7. What is an Iterator class?
Answer:
A
class that is used to traverse through the objects maintained by a container
class. There are five categories of iterators:
Ø input iterators,
Ø output
iterators,
Ø forward
iterators,
Ø bidirectional
iterators,
Ø random access.
An iterator is
an entity that gives access to the contents of a container object without
violating encapsulation constraints. Access to the contents is granted on a
one-at-a-time basis in order. The order can be storage order (as in lists and
queues) or some arbitrary order (as in array indices) or according to some
ordering relation (as in an ordered binary tree). The iterator is a construct,
which provides an interface that, when called, yields either the next element
in the container, or some value denoting the fact that there are no more
elements to examine. Iterators hide the details of access to and update of the
elements of a container class.
The simplest and
safest iterators are those that permit read-only access to the contents of a
container class. The following code fragment shows how an iterator might appear
in code:
cont_iter:=new cont_iterator();
x:=cont_iter.next();
while x/=none do
...
s(x);
...
x:=cont_iter.next();
end;
In this example, cont_iter is the name
of the iterator. It is created on the first line by instantiation of
cont_iterator class, an iterator class defined to iterate over some container
class, cont. Succesive elements from the container are carried to x. The loop
terminates when x is bound to some empty value. (Here, none)In the middle of
the loop, there is s(x) an operation on x, the current element from the
container. The next element of the container is obtained at the bottom of the
loop.
9. List out some of the OODBMS
available.
Answer:
Ø GEMSTONE/OPAL of Gemstone systems.
Ø ONTOS of Ontos.
Ø Objectivity of Objectivity inc.
Ø Versant of Versant object technology.
Ø Object store of Object Design.
Ø ARDENT of ARDENT software.
Ø POET of POET software.
10. List out some of the
object-oriented methodologies.
Answer:
Ø Object Oriented Development (OOD) (Booch 1991,1994).
Ø Object Oriented Analysis and Design (OOA/D) (Coad and Yourdon 1991).
Ø Object Modelling Techniques (OMT)
(Rumbaugh 1991).
Ø Object Oriented Software Engineering (Objectory) (Jacobson 1992).
Ø Object Oriented Analysis (OOA) (Shlaer and Mellor 1992).
Ø The Fusion Method (Coleman 1991).
11. What is an incomplete type?
Answer:
Incomplete
types refers to pointers in which there is non availability of the
implementation of the referenced location or it points to some location whose
value is not available for modification.
Example:
int *i=0x400 // i points to address 400
*i=0; //set the value of memory location
pointed by i.
Incomplete types are otherwise
called uninitialized pointers.
12. What is a dangling pointer?
Answer:
A dangling
pointer arises when you use the address of an object after its lifetime is
over.
This may occur in situations like
returning addresses of the automatic variables from a function or using the
address of the memory block after it is freed.
13. Differentiate between the
message and method.
Answer:
Message
Method
Objects communicate by sending
messages Provides response to a
message.
to each other.
A message is sent to invoke a
method. It is an
implementation of an operation.
14. What is an adaptor class or
Wrapper class?
Answer:
A class that has
no functionality of its own. Its member functions hide the use of a third party
software component or an object with the non-compatible interface or a non-
object- oriented implementation.
15. What is a Null object?
Answer:
It is an object
of some class whose purpose is to indicate that a real object of that class
does not exist. One common use for a null object is a return value from a
member function that is supposed to return an object with some specified
properties but cannot find such an object.
16. What is class invariant?
Answer:
A class
invariant is a condition that defines all valid states for an object. It is a
logical condition to ensure the correct working of a class. Class invariants
must hold when an object is created, and they must be preserved under all
operations of the class. In particular all class invariants are both
preconditions and post-conditions for all operations or member functions of the
class.
17. What do you mean by Stack
unwinding?
Answer:
It is a process
during exception handling when the destructor is called for all local objects
between the place where the exception was thrown and where it is caught.
18. Define precondition and
post-condition to a member function.
Answer:
Precondition:
A precondition is a condition that
must be true on entry to a member function. A class is used correctly if
preconditions are never false. An operation is not responsible for doing
anything sensible if its precondition fails to hold.
For example, the
interface invariants of stack class say
nothing about pushing yet another element on a stack that is already full. We
say that isful() is a precondition of
the push operation.
Post-condition:
A post-condition is a condition
that must be true on exit from a member function if the precondition was valid
on entry to that function. A class is implemented correctly if post-conditions
are never false.
For example,
after pushing an element on the stack, we know that isempty() must necessarily hold. This is a post-condition of the push operation.
19. What are the conditions that have to be met
for a condition to be an invariant of the class?
Answer:
Ø The
condition should hold at the end of every constructor.
Ø The
condition should hold at the end of every mutator(non-const) operation.
20. What are proxy objects?
Answer:
Objects that stand for other
objects are called proxy objects or surrogates.
Example:
template<class T>
class Array2D
{
public:
class Array1D
{
public:
T&
operator[] (int index);
const T&
operator[] (int index) const;
...
};
Array1D operator[]
(int index);
const Array1D
operator[] (int index) const;
...
};
The following then becomes legal:
Array2D<float>data(10,20);
........
cout<<data[3][6]; //
fine
Here
data[3] yields an Array1D object and the operator [] invocation on that object
yields the float in position(3,6) of the original two dimensional array.
Clients of the Array2D class need not be aware of the presence of the Array1D
class. Objects of this latter class stand for one-dimensional array objects
that, conceptually, do not exist for clients of Array2D. Such clients program
as if they were using real, live, two-dimensional arrays. Each Array1D object
stands for a one-dimensional array that is absent from a conceptual model used
by the clients of Array2D. In the above example, Array1D is a proxy class. Its
instances stand for one-dimensional arrays that, conceptually, do not exist.
21. Name some pure object oriented languages.
Answer:
Ø Smalltalk,
Ø Java,
Ø Eiffel,
Ø Sather.
22. Name the operators that cannot be
overloaded.
Answer:
sizeof . .* .-> :: ?:
23. What is a node class?
Answer:
A node class is
a class that,
Ø relies
on the base class for services and implementation,
Ø provides
a wider interface to te users than its base class,
Ø relies
primarily on virtual functions in its public interface
Ø depends
on all its direct and indirect base class
Ø can
be understood only in the context of the base class
Ø can
be used as base for further derivation
Ø can
be used to create objects.
A node class is a class that has
added new services or functionality beyond the services inherited from its base
class.
24.
What is an orthogonal base class?
Answer:
If two base classes have no overlapping
methods or data they are said to be independent of, or orthogonal to each
other. Orthogonal in the sense means that two classes operate in different
dimensions and do not interfere with each other in any way. The same derived
class may inherit such classes with no difficulty.
25. What is a container class? What are the
types of container classes?
Answer:
A container
class is a class that is used to hold objects in memory or external storage. A
container class acts as a generic holder. A container class has a predefined
behavior and a well-known interface. A container class is a supporting class
whose purpose is to hide the topology used for maintaining the list of objects
in memory. When a container class contains a group of mixed objects, the
container is called a heterogeneous container; when the container is holding a
group of objects that are all the same, the container is called a homogeneous
container.
26. What is a protocol class?
Answer:
An abstract
class is a protocol class if:
Ø it
neither contains nor inherits from classes that contain member data,
non-virtual functions, or private (or protected) members of any kind.
Ø it
has a non-inline virtual destructor defined with an empty implementation,
Ø all
member functions other than the destructor including inherited functions, are
declared pure virtual functions and left undefined.
27. What is a mixin class?
Answer:
A class that
provides some but not all of the implementation for a virtual base class is
often called mixin. Derivation done just for the purpose of redefining the
virtual functions in the base classes is often called mixin inheritance. Mixin
classes typically don't share common bases.
28. What is a concrete class?
Answer:
A concrete class
is used to define a useful object that can be instantiated as an automatic
variable on the program stack. The implementation of a concrete class is
defined. The concrete class is not intended to be a base class and no attempt
to minimize dependency on other classes in the implementation or behavior of
the class.
29.What is the handle class?
Answer:
A handle is a
class that maintains a pointer to an object that is programmatically accessible
through the public interface of the handle class.
Explanation:
In case of abstract classes, unless one
manipulates the objects of these classes through pointers and references, the
benefits of the virtual functions are lost. User code may become dependent on
details of implementation classes because an abstract type cannot be allocated
statistically or on the stack without its size being known. Using pointers or
references implies that the burden of memory management falls on the user.
Another limitation of abstract class object is of fixed size. Classes however
are used to represent concepts that require varying amounts of storage to
implement them.
A popular technique for dealing
with these issues is to separate what is used as a single object in two parts:
a handle providing the user interface and a representation holding all or most
of the object's state. The connection between the handle and the representation
is typically a pointer in the handle. Often, handles have a bit more data than
the simple representation pointer, but not much more. Hence the layout of the
handle is typically stable, even when the representation changes and also that
handles are small enough to move around relatively freely so that the user
needn’t use the pointers and the references.
30.
What is an action class?
Answer:
The simplest and
most obvious way to specify an action in C++ is to write a function. However,
if the action has to be delayed, has to be transmitted 'elsewhere' before being
performed, requires its own data, has to be combined with other actions, etc
then it often becomes attractive to provide the action in the form of a class
that can execute the desired action and provide other services as well.
Manipulators used with iostreams is an obvious example.
Explanation:
A
common form of action class is a simple class containing just one virtual
function.
class Action
{
public:
virtual int do_it( int )=0;
virtual ~Action( );
}
Given this, we can write code say a member
that can store actions for later execution without using pointers to functions,
without knowing anything about the objects involved, and without even knowing
the name of the operation it invokes. For example:
class write_file
: public Action
{
File& f;
public:
int do_it(int)
{
return fwrite( ).suceed(
);
}
};
class error_message: public Action
{
response_box db(message.cstr(
),"Continue","Cancel","Retry");
switch (db.getresponse( ))
{
case 0: return 0;
case 1: abort();
case 2:
current_operation.redo( );return 1;
}
};
A user of the Action class will be completely isolated from any
knowledge of derived classes such as write_file and error_message.
31. When can you tell that a memory leak
will occur?
Answer:
A memory leak
occurs when a program loses the ability to free a block of dynamically
allocated memory.
32.What is a parameterized type?
Answer:
A template is a
parameterized construct or type containing generic code that can use or
manipulate any type. It is called parameterized because an actual type is a
parameter of the code body. Polymorphism may be achieved through parameterized
types. This type of polymorphism is called parameteric polymorphism.
Parameteric polymorphism is the mechanism by which the same code is used on
different types passed as parameters.
33. Differentiate between a deep copy and a
shallow copy?
Answer:
Deep copy
involves using the contents of one object to create another instance of the
same class. In a deep copy, the two objects may contain ht same information but
the target object will have its own buffers and resources. the destruction of
either object will not affect the remaining object. The overloaded assignment
operator would create a deep copy of objects.
Shallow copy
involves copying the contents of one object into another instance of the same
class thus creating a mirror image. Owing to straight copying of references and
pointers, the two objects will share the same externally contained contents of
the other object to be unpredictable.
Explanation:
Using a copy
constructor we simply copy the data values member by member. This method of
copying is called shallow copy. If the object is a simple class, comprised of
built in types and no pointers this would be acceptable. This function would
use the values and the objects and its behavior would not be altered with a
shallow copy, only the addresses of pointers that are members are copied and
not the value the address is pointing to. The data values of the object would
then be inadvertently altered by the function. When the function goes out of
scope, the copy of the object with all its data is popped off the stack.
If the object
has any pointers a deep copy needs to be executed. With the deep copy of an
object, memory is allocated for the object in free store and the elements
pointed to are copied. A deep copy is used for objects that are returned from a
function.
34. What is an opaque pointer?
Answer:
A pointer is
said to be opaque if the definition of the type to which it points to is not
included in the current translation unit. A translation unit is the result of
merging an implementation file with all its headers and header files.
35. What is a smart pointer?
Answer:
A
smart pointer is an object that acts, looks and feels like a normal pointer but
offers more functionality. In C++, smart pointers are implemented as template classes that encapsulate a
pointer and override standard pointer operators. They have a number of
advantages over regular pointers. They are guaranteed to be initialized as
either null pointers or pointers to a heap object. Indirection through a null
pointer is checked. No delete is ever necessary. Objects are automatically
freed when the last pointer to them has gone away. One significant problem with
these smart pointers is that unlike regular pointers, they don't respect
inheritance. Smart pointers are unattractive for polymorphic code. Given below
is an example for the implementation of smart pointers.
Example:
template <class X>
class smart_pointer
{
public:
smart_pointer(); // makes a null
pointer
smart_pointer(const X&
x) // makes pointer to copy of
x
X& operator *( );
const X& operator*( )
const;
X* operator->() const;
smart_pointer(const
smart_pointer <X> &);
const smart_pointer
<X> & operator =(const smart_pointer<X>&);
~smart_pointer();
private:
//...
};
This
class implement a smart pointer to an object of type X. The object itself is
located on the heap. Here is how to use it:
smart_pointer <employee> p=
employee("Harris",1333);
Like
other overloaded operators, p will behave like a regular pointer,
cout<<*p;
p->raise_salary(0.5);
36. What is reflexive association?
Answer:
The
'is-a' is called a reflexive association because the reflexive association
permits classes to bear the is-a association not only with their super-classes
but also with themselves. It differs from a 'specializes-from' as 'specializes-from' is usually used to
describe the association between a super-class and a sub-class. For example:
Printer
is-a printer.
37.
What is slicing?
Answer:
Slicing
means that the data added by a subclass are discarded when an object of the
subclass is passed or returned by value or from a function expecting a base
class object.
Explanation:
Consider
the following class declaration:
class base
{
...
base& operator =(const
base&);
base (const base&);
}
void fun( )
{
base e=m;
e=m;
}
As
base copy functions don't know anything about the derived only the base part of
the derived is copied. This is commonly referred to as slicing. One reason to
pass objects of classes in a hierarchy is to avoid slicing. Other reasons are
to preserve polymorphic behavior and to gain efficiency.
38. What is name mangling?
Answer:
Name mangling is the process through which
your c++ compilers give each function in your program a unique name. In C++,
all programs have at-least a few functions with the same name. Name mangling is
a concession to the fact that linker always insists on all function names being
unique.
Example:
In general, member names are made
unique by concatenating the name of the member with that of the class e.g.
given the declaration:
class Bar
{
public:
int ival;
...
};
ival becomes
something like:
// a possible member name mangling
ival__3Bar
Consider this
derivation:
class Foo : public Bar
{
public:
int ival;
...
}
The internal representation of a Foo object
is the concatenation of its base and derived class members.
// Pseudo C++ code
// Internal representation of Foo
class Foo
{
public:
int ival__3Bar;
int ival__3Foo;
...
};
Unambiguous access of either ival members is
achieved through name mangling. Member functions, because they can be
overloaded, require an extensive mangling to provide each with a unique name.
Here the compiler generates the same name for the two overloaded
instances(Their argument lists make their instances unique).
39. What are proxy objects?
Answer:
Objects
that points to other objects are called proxy objects or surrogates. Its an
object that provides the same interface as its server object but does not have
any functionality. During a method invocation, it routes data to the true
server object and sends back the return value to the object.
40. Differentiate between declaration and
definition in C++.
Answer:
A declaration introduces a name into the
program; a definition provides a unique description of an entity (e.g. type,
instance, and function). Declarations can be repeated in a given scope, it
introduces a name in a given scope. There must be exactly one definition of
every object, function or class used in a C++ program.
A
declaration is a definition unless:
Ø it
declares a function without specifying its body,
Ø it
contains an extern specifier and no initializer or function body,
Ø it
is the declaration of a static class data member without a class definition,
Ø it
is a class name definition,
Ø it
is a typedef declaration.
A
definition is a declaration unless:
Ø it
defines a static class data member,
Ø it
defines a non-inline member function.
41. What is cloning?
Answer: An object can carry out copying
in two ways i.e. it can set itself to be a copy of another object, or it can
return a copy of itself. The latter process is called cloning.
42. Describe the main characteristics of
static functions.
Answer:
The main characteristics of static
functions include,
Ø It
is without the a this pointer,
Ø It
can't directly access the non-static members of its class
Ø It
can't be declared const, volatile or virtual.
Ø It
doesn't need to be invoked through an object of its class, although for
convenience, it may.
43. Will the inline function be compiled as
the inline function always? Justify.
Answer:
An inline function is a request and
not a command. Hence it won't be compiled as an inline function always.
Explanation:
Inline-expansion could fail if the
inline function contains loops, the address of an inline function is used, or
an inline function is called in a complex expression. The rules for inlining
are compiler dependent.
44. Define a way other than using the
keyword inline to make a function inline.
Answer:
The function must be defined inside
the class.
45. How can a '::' operator be used as unary
operator?
Answer:
The scope operator can be used to
refer to members of the global namespace. Because the global namespace doesn’t
have a name, the notation :: member-name refers to a member of the global
namespace. This can be useful for referring to members of global namespace
whose names have been hidden by names declared in nested local scope. Unless we
specify to the compiler in which namespace to search for a declaration, the
compiler simple searches the current scope, and any scopes in which the current
scope is nested, to find the declaration for the name.
46. What is placement new?
Answer:
When
you want to call a constructor directly, you use the placement new. Sometimes
you have some raw memory that's already been allocated, and you need to
construct an object in the memory you have. Operator new's special version placement
new allows you to do it.
class Widget
{
public :
Widget(int widgetsize);
...
Widget*
Construct_widget_int_buffer(void *buffer,int widgetsize)
{
return
new(buffer) Widget(widgetsize);
}
};
This
function returns a pointer to a Widget object that's constructed within the
buffer passed to the function. Such a function might be useful for applications
using shared memory or memory-mapped I/O, because objects in such applications
must be placed at specific addresses or in memory allocated by special
routines.
OOAD
1.
What
do you mean by analysis and design?
Analysis:
Basically, it is the process of
determining what needs to be done before how it should be done. In order to
accomplish this, the developer refers the existing systems and documents. So,
simply it is an art of discovery.
Design:
It is the process of adopting/choosing
the one among the many, which best accomplishes the users needs. So, simply, it
is compromising mechanism.
2.
What
are the steps involved in designing?
Before getting into the design the
designer should go through the SRS prepared by the System Analyst.
The
main tasks of design are Architectural Design and Detailed Design.
In
Architectural Design we find what are the main modules in the problem domain.
In Detailed Design we find what
should be done within each module.
3.
What
are the main underlying concepts of object orientation?
Objects, messages, class, inheritance and
polymorphism are the main concepts of object orientation.
4.
What
do u meant by "SBI" of an object?
SBI stands for State, Behavior and
Identity. Since every object has the above three.
Ø
State:
It is just a value to
the attribute of an object at a particular time.
Ø
Behaviour:
It describes the
actions and their reactions of that object.
Ø
Identity:
An object has an identity that
characterizes its own existence. The identity makes it possible to distinguish
any object in an unambiguous way, and independently from its state.
5.
Differentiate
persistent & non-persistent objects?
Persistent refers to an object's
ability to transcend time or space. A persistent object stores/saves its state
in a permanent storage system with out losing the information represented by
the object.
A non-persistent object is said to be
transient or ephemeral. By default objects are considered as non-persistent.
6.
What
do you meant by active and passive objects?
Active objects are one which
instigate an interaction which owns a thread and they are responsible for
handling control to other objects. In simple words it can be referred as client.
Passive objects are one, which
passively waits for the message to be processed. It waits for another object
that requires its services. In simple words it can be referred as server.
Diagram:
client server
(Active) (Passive)
7.
What
is meant by software development method?
Software development method describes
how to model and build software systems in a reliable and reproducible way. To
put it simple, methods that are used to represent ones' thinking using
graphical notations.
8.
What
are models and meta models?
Model:
It is a complete description of something (i.e. system).
Meta
model:
It describes the model elements,
syntax and semantics of the notation that allows their manipulation.
9.
What
do you meant by static and dynamic modeling?
Static modeling is used to specify
structure of the objects that exist in the problem domain. These are expressed
using class, object and USECASE diagrams.
But Dynamic modeling refers
representing the object interactions during runtime. It is represented by sequence, activity, collaboration and
statechart diagrams.
10. How to represent the interaction between the
modeling elements?
Model
element is just a notation to represent (Graphically) the entities that
exist in the problem domain. e.g. for modeling element is class notation,
object notation etc.
Relationships are used to represent the
interaction between the modeling elements.
The following are the Relationships.
Ø
Association: Its' just a semantic connection two classes.
e.g.:
Ø
Aggregation: Its' the relationship between two classes which are related in the
fashion that master and slave. The
master takes full rights than the slave. Since the slave works under the
master. It is represented as line with diamond in the master area.
ex:
car contains wheels, etc.
car
Ø
Containment: This relationship is applied when the part contained with in the whole
part, dies when the whole part dies.
It is represented as darked diamond at the
whole part.
example:
class A{
//some code
};
class B
{
A aa; // an
object of class A;
// some
code for class B;
};
In the above example we see that an object of
class A is instantiated with in the class
B. so the object class A dies when the object class B dies.we can
represnt it in diagram like this.
Ø
Generalization: This relationship used when we want represents a
class, which captures the common states of objects of different classes. It is
represented as arrow line pointed at the class, which has captured the common
states.
Ø
Dependency: It is the relationship between dependent and independent classes. Any
change in the independent class will affect the states of the dependent class.
DIAGRAM:
class
A class B
11. Why
generalization is very strong?
Even though Generalization satisfies
Structural, Interface, Behaviour properties. It is mathematically very strong,
as it is Antisymmetric and Transitive.
Antisymmetric: employee is a
person, but not all persons are employees. Mathematically all As’ are B, but
all Bs’ not A.
Transitive: A=>B, B=>c then A=>c.
A. Salesman.
B. Employee.
C. Person.
Note: All the other relationships satisfy all
the properties like Structural properties, Interface properties, Behaviour
properties.
12. Differentiate
Aggregation and containment?
Aggregation
is the relationship between the whole and a part. We can add/subtract some properties in the part (slave) side. It won't
affect the whole part.
Best
example is Car, which contains the wheels and some extra parts. Even though the
parts are not there we can call it as car.
But,
in the case of containment the whole part is affected when the part within that
got affected. The human body is an apt example for this relationship. When the
whole body dies the parts (heart etc) are died.
13. Can link
and Association applied interchangeably?
No,
You cannot apply the link and Association interchangeably. Since link is used
represent the relationship between the two objects.
But
Association is used represent the relationship between the two classes.
link
:: student:Abhilash course:MCA
Association:: student course
14. what is
meant by "method-wars"?
Before 1994 there were different
methodologies like Rumbaugh, Booch, Jacobson, Meyer etc who followed their own
notations to model the systems. The developers were in a dilemma to choose the
method which best accomplishes their needs.
This particular span was called as "method-wars"
15. Whether
unified method and unified modeling language are same or different?
Unified method is convergence of the Rumbaugh and Booch.
Unified modeling lang. is the fusion of Rumbaugh, Booch and Jacobson as
well as Betrand Meyer (whose contribution is "sequence diagram").
Its' the superset of all the methodologies.
16. Who were
the three famous amigos and what was their contribution to the object community?
The Three amigos namely,
Ø
James Rumbaugh (OMT): A veteran in analysis who came up with an idea
about the objects and their
Relationships (in particular Associations).
Ø
Grady Booch: A veteran in design who came up with an idea about partitioning of
systems into subsystems.
Ø
Ivar Jacobson (Objectory): The father of USECASES, who described about the
user and system interaction.
17. Differentiate
the class representation of Booch, Rumbaugh and UML?
If you look at the class representaiton of
Rumbaugh and UML, It is some what similar and both are very easy to draw.
Representation: OMT UML.
Diagram:
Booch:
In this method classes are represented as "Clouds" which are not very
easy to draw as for as the developer's view is concern.
Diagram:
18. What is
an USECASE? Why it is needed?
A
Use Case is a description of a set of sequence of actions that a system
performs that yields an observable result of value to a particular action.
In SSAD process <=> In OOAD USECASE. It is
represented elliptically.
Representation:
19. Who is
an Actor?
An
Actor is someone or something that must interact with the system.In addition to
that an Actor initiates the process(that
is USECASE).
It
is represented as a stickman like this.
Diagram:
 |
20. What is
guard condition?
Guard
condition is one, which acts as a firewall. The access from a particular object
can be made only when the particular condition is met.
For
Example,
customer check customer number ATM.
Here the object on the customer accesses the ATM
facility only when the guard condition is met.
21. Differentiate
the following notations?
I: :obj1 :obj2
II: :obj1 :obj2
In
the above representation I, obj1 sends message to obj2. But in the case of II
the data is transferred from obj1 to obj2.
22. USECASE
is an implementation independent notation. How will the designer give the
implementation details of a particular USECASE to the programmer?
This can be accomplished by
specifying the relationship called "refinement” which talks about the two different
abstraction of the same thing.
Or
example,
calculate
pay calculate
class1 class2 class3
23. Suppose
a class acts an Actor in the problem domain, how to represent it in the static
model?
In
this scenario you can use “stereotype”. Since stereotype is just a string that
gives extra semantic to the particular entity/model element. It is given with
in the << >>.
class
A
<<
Actor>>
attributes
methods.
24. Why does
the function arguments are called as "signatures"?
The
arguments distinguish functions with the same name (functional polymorphism).
The name alone does not necessarily identify a unique function. However, the name and its arguments
(signatures) will uniquely identify a function.
In
real life we see suppose, in class there are two guys with same name, but they
can be easily identified by their
signatures. The same concept is applied here.
ex:
class
person
{
public:
char
getsex();
void
setsex(char);
void
setsex(int);
};
In
the above example we see that there is a function setsex() with same name but
with different signature.
Playing with scanf function
Operators & Expressions
_____ _____________________________________
[Q001]. Determine which of the following are VALID
identifiers. If invalid, state the reason.
(a) sample1 (b) 5sample (c) data_7 (d)
return (e) #fine
(f)
variable (g) 91-080-100 (h) name & age (i) _val (j) name_and_age
Ans. (a) VALID
(b)
Invalid, since an identifier must begin with a letter or an underscore
(c) VALID
(d)
Invalid, since return is a reserved word
(e)
Invalid, since an identifier must begin with a letter or an underscore
(f) VALID
(g)
Invalid, since an identifier must begin with a letter or an underscore
(h)
Invalid, since blank spaces are not allowed
(i) VALID
(j) VALID
_________________________________________________________________________________________________
[Q002]. Determine which of the following are VALID character
constants. If invalid, state the reason.
(a) 'y' (b) '\r' (c) 'Y' (d) '@' (e) '/r'
(f) 'word' (g) '\0' (h) '\?' (i) '\065' (j) '\'' (k) ' '
Ans. (a) VALID
(b) VALID
(c) VALID
(d) VALID
(e)
Invalid, since escape sequences must be written with a backward slash (i.e. \)
(f)
Invalid, since a character constant cannot consist of multiple characters
(g) VALID
(null-character escape sequence)
(h) VALID
(i) VALID
(Octal escape sequence)
(j) VALID
(k) VALID
_________________________________________________________________________________________________
[Q003]. Determine which of the following are VALID string
constants. If invalid, state the reason.
(a) 'Hi
Friends' (b)
"abc,def,ghi" (c)
"Qualification
(d)
"4325.76e-8" (e)
"Don\'t sleep" (f)
"He said, "You\'re great"
(g)
"" (h)
" " (i) "Rs.100/-"
Ans. (a) Invalid,
since a string constant must be enclosed in double quotation marks
(b) VALID
(c)
Invalid, since trailing quotation mark is missing
(d) VALID
(e) VALID
(single-quote escape sequence)
(f)
Invalid, since the quotation marks and (optionally) apostrophe within the
string
cannot be expressed without the escape
sequences.
(g) VALID
(h) VALID
(i) VALID
_________________________________________________________________________________________________
[Q004]. Determine which of the following numerical values
are valid constants. If a constant is
valid, specify whether it is integer or real. Also, specify
the base for each valid integer constant.
(a) 10,500 (b) 080 (c)
0.007 (d) 5.6e7 (e) 5.6e-7
(f)
0.2e-0.3 (g) 0.2e 0.3 (h) 0xaf9s82 (i) 0XABCDEFL (j)
0369CF
(k)
87654321l (l) 87654321
Ans. (a) Invalid,
since illegal character(,)
(b) VALID
(c) VALID
(d) VALID
(e) VALID
(f) VALID
(g)
Invalid, since illegal character(blank space)
(h)
Invalid, since illegal character(s)
(i) VALID
(j)
Invalid, since illegal characters (9, C, F), if intended as an octal constant.
(k) VALID
(l) VALID
_________________________________________________________________________________________________
[Q005]. Determine which of the following floating-point
constants are VALID for the quantity (5 * 100000).
(a) 500000 (b) 0.5e6 (c)
5E5 (d) 5e5 (e) 5e+5
(f) 500E3 (g) .5E6 (h) 50e4 (i)
50.E+4 (j) 5.0E+5
(k) All of
the above (l) None of these
Ans. (k)
_________________________________________________________________________________________________
[Q006]. What will be the output of the following program :
void main()
{
printf("%f",123.);
}
(a)123 (b)Compile-Time
Error (c)123.00 (d)123.000000
Ans. (d)
_________________________________________________________________________________________________
[Q007]. What will be the output of the following program :
void main()
{
printf("%d",sizeof(integer));
}
(a)2 (b)Compile-Time
Error (c)4 (d)None of these
Ans. (b) since there is no such data type called 'integer'.
_________________________________________________________________________________________________
[Q008]. What will be the output of the following program :
void main()
{
char
str[]="C For Swimmers";
printf("%d",sizeof str);
}
(a)14 (b)Compile-Time
Error (c)15 (d)None of these
Ans. (a)--c
_________________________________________________________________________________________________
[Q009]. What will be the output of the following program :
void main()
{
char
str[]="C For Swimmers";
printf("%d",++(sizeof(str)));
}
(a)14 (b)Compile-Time
Error (c)15 (d)None of these
Ans. (b)
_________________________________________________________________________________________________
[Q010]. What will be the output of the following program :
void main()
{
char
str[]="C For Swimmers";
printf("%d",-sizeof(str));
}
(a)14 (b)Compile-Time
Error (c)-15 (d)-14
Ans. (c)
_________________________________________________________________________________________________
[Q011]. What will be the output of the following program :
void
main()
{
printf("%d",!(100==100)+1);
}
(a)100 (b)0 (c)1 (d)2
Ans. (c)
_________________________________________________________________________________________________
[Q012]. What will be the output of the following program :
void main()
{
int
x=5,y=6,z=2;
z/=y/z==3?y/z:x*y;
printf("%d",z);
}
(a)Compile-Time Error (b)2 (c)0 (d)1
Ans. (c)
_________________________________________________________________________________________________
[Q013]. What will be the output of the following program :
void main()
{
printf("%d %d %d",5,!5,25 - !25);
}
(a)5 10 22 (b)5
5 25 (c)5
0 25 (d)5 1 24
Ans. (c)
_________________________________________________________________________________________________
[Q014]. What will be the output of the following program :
int main()
{
int a=500,b=100,c=30,d=40,e=19;
a+=b-=c*=d/=e%=5;
printf("%d %d %d %d %d",a,b,c,d,e);
}
(a)500 100 30 40 4 (b)Run-Time
Error (c)700 200 300 10 4 (d)300 -200 300 10 4
Ans. (d)
_________________________________________________________________________________________________
[Q015]. What will be the output of the following program :
void main()
{
int
a=500,b=100,c=30,d=40,e=19;
if ((((a > b) ? c : d) >= e) &&
!((e <= d) ? ((a / 5) == b) : (c == d)))
printf("Success");
else
printf("Failure");
}
(a)VALID : Success (b)VALID
: Failure (c)INVALID (d)None of these
Ans. (b)
_________________________________________________________________________________________________
[Q016]. What will be the output of the following program :
void main()
{
int
a=1,b=2,c=3,d=4;
printf("%d",!a?b?!c:!d:a);
}
(a)1 (b)2 (c)3 (d)4
Ans. (a)
_________________________________________________________________________________________________
[Q017]. What will be the output of the following program :
void main()
{
int i=12345,j=-13579,k=-24680;
long ix=123456789;
short sx=-2222;
unsigned ux=5555;
printf("\n%d %d %d %ld %d
%u",i,j,k,ix,sx,ux);
printf("\n\n%3d %3d %3d\n%3ld %3d
%3u",i,j,k,ix,sx,ux);
printf("\n\n%8d %8d %8d\n%15ld %8d
%8u",i,j,k,ix,sx,ux);
printf("\n\n%-8d %-8d\n%-8d %-15ld\n%-8d
%-8u",i,j,k,ix,sx,ux);
printf("\n\n%+8d %+8d\n%+8d %+15ld\n%+8d
%8u",i,j,k,ix,sx,ux);
printf("\n\n%08d %08d\n%08d %015ld\n%08d
%08u",i,j,k,ix,sx,ux);
}
Ans. 12345 -13579 -24680 123456789 -2222 5555
12345 -13579
-24680
123456789 -2222
5555
12345 -13579
-24680
123456789 -2222 5555
12345 -13579
-24680 123456789
-2222 5555
+12345 -13579
-24680 +123456789
-2222 5555
00012345 -0013579
-0024680
000000123456789
-0002222 00005555
_________________________________________________________________________________________________
[Q018]. What will be the output of the following program :
void main()
{
int i=12345,j=0xabcd9,k=077777;
printf("%d %x %o",i,j,k);
printf("\n%3d %3x %3o",i,j,k);
printf("\n%8d %8x %8o"i,j,k);
printf("\n%-8d %-8x %-8o",i,j,k);
printf("\n%+8d %+8x %+8o",i,j,k);
printf("\n%08d %#8x %#8o",i,j,k);
}
Ans. 12345 abcd9 77777
12345 abcd9 77777
12345 abcd9
77777
12345 abcd9
77777
+12345 abcd9
77777
00012345 0xabcd9
077777
_________________________________________________________________________________________________
[Q019]. What will be the output of the following program :
void main()
{
char c1='A', c2='B', c3='C';
printf("%c %c %c",c1,c2,c3);
printf("\n%c%c%c",c1,c2,c3);
printf("\n%3c %3c %3c",c1,c2,c3);
printf("\n%3c%3c%3c",c1,c2,c3);
printf("\nc1=%c c2=%c
c3=%c",c1,c2,c3);
}
Ans. A B C
ABC
A B C
A B C
c1=A c2=B c3=C
_________________________________________________________________________________________________
[Q020]. What will be the output of the following program :
void main()
{
float a=2.5, b=0.0005, c=3000.;
printf("%f %f %f",a,b,c);
printf("\n%3f %3f %3f",a,b,c);
printf("\n%8f %8f %8f",a,b,c);
printf("\n%8.4f %8.4f
%8.4f",a,b,c);
printf("\n%8.3f %8.3f
%8.3f",a,b,c);
printf("\n%e %e %e",a,b,c);
printf("\n%3e %3e %3e",a,b,c);
printf("\n%12e %12e %12e",a,b,c);
printf("\n%8.2e %8.2e
%8.2e",a,b,c);
printf("\n%-8f %-8f %-8f",a,b,c);
printf("\n%+8f %+8f %+8f",a,b,c);
printf("\n%08f %08f %08f",a,b,c);
printf("\n%#8f %#8f %#8f",a,b,c);
printf("\n%g %g %g",a,b,c);
printf("\n%#g %#g %#g"a,b,c);
}
Ans. 2.500000 0.000500 3000.000000
2.500000 0.000500
3000.000000
2.500000 0.000500
3000.000000
2.5000 0.0005 3000.0000
2.500 0.001 3000.000
2.500000e+000
5.000000e-004 3.000000e+003
2.500000e+000
5.000000e-004 3.000000e+003
2.500000e+000
5.000000e-004 3.000000e+003
2.5000e+000 5.0000e-004
3.0000e+003
2.50e+000
5.00e-004 3.00e+003
2.500000 0.000500
3000.000000
+2.500000
+0.000500 +3000.000000
2.500000 0.000500
3000.000000
2.500000 0.000500
3000.000000
2.5 0.0005 3000
2.500000 0.000500
3000.000000
_________________________________________________________________________________________________
[Q021]. What will be the output of the following program :
void main()
{
char str[]="C For Swimmers";
printf("%s",str);
printf("\n%.5s",str);
printf("\n%8.*s",5,str);
printf("\n%-10s %.1s",str+6,str);
}
Ans. C For Swimmers
C For
C For
Swimmers C
_________________________________________________________________________________________________
[Q022]. What will be the output of the following program :
void
main()
{
int a=1,b=2,c=3;
scanf("%d %*d
%d",&a,&b,&c);
printf("a=%d b=%d c=%d",a,b,c);
}
[NOTE : 3 values entered by the user are:100 200 300]
(a)1 2 3 (b)100
200 300 (c)100
200 3 (d)100 300 3
Ans. (d)
_________________________________________________________________________________________________
[Q023]. What will be the output of the following program :
void main()
{
char line[80]; // Max. length=80 Chars
scanf("%[^,]s",line);
printf("\n%s",line);
}
[NOTE : THE USER INPUT IS:Dear Friends, What is the output?]
(a)Compile-Time Error (b)Dear
Friends (c)What
is the output? (d)None of these
Ans. (b)
_________________________________________________________________________________________________
[Q024]. What will be the output of the following program :
void main()
{
char a,b,c;
scanf("%c%c%c",&a,&b,&c);
printf("a=%c b=%c c=%c",a,b,c);
}
[NOTE : THE USER INPUT IS :A B C]
(a)a=A b=B c=C (b)a=A
b= c=B (c)a=A
b= c=C (d)None
of these
Ans. (b)
_________________________________________________________________________________________________
[Q025]. What will be the output of the following program :
void main()
{
int i=1;
float f=2.25;
scanf("%d a %f",&i,&f);
printf("%d %.2f",i,f);
}
[NOTE : THE USER INPUT IS:5 5.75]
(a)1 2.25 (b)5
5.75 (c)5
2.25 (d)None of these
Ans. (c)
_________________________________________________________________________________________________
[Q026]. What will be the output of the following program :
void main()
{
char a,b,c;
scanf("%c %c
%c",&a,&b,&c);
printf("a=%c b=%c c=%c",a,b,c);
}
[NOTE : THE USER INPUT IS :ABC DEF GHI]
(a)a=ABC b=DEF c=GHI (b)a=A
b=B c=C (c)a=A b=D c=G (d)None of these
Ans. (b)
_________________________________________________________________________________________________
[Q027]. What will be the output of the following program :
void main()
{
char a[80],b[80],c[80];
scanf("%1s %5s %3s",a,b,c);
printf("%s %s %s",a,b,c);
}
[NOTE : THE USER INPUT IS:CMeansSea Ocean Vast]
(a)C O V (b)C
Means Sea (c)C
Ocean Vas (d)None of
these
Ans. (b)
_________________________________________________________________________________________________
[Q028]. What will be the output of the following program :
void main()
{
int a,b,c;
scanf("%1d %2d
%3d",&a,&b,&c);
printf("Sum=%d",a+b+c);
}
[NOTE
: THE USER INPUT IS :123456 44 544]
(a)Sum=480 (b)Sum=594 (c)Sum=589 (d)None of these
Ans. (a)
_________________________________________________________________________________________________
[Q029]. What happens when the following program is executed
:
void main()
{
char line[80];
scanf("%[^1234567890\n]",line);
}
(a)Accepts the string that contains DIGITS only.
(b)Accepts the string that contains DIGITS and NEWLINE
characters.
(c)Accepts the string that contains anything other than the
DIGITS and NEWLINE characters.
(d)None of these
Ans. (c)
_________________________________________________________________________________________________
[Q030]. What happens when the following program is executed
:
void main()
{
char line[80];
scanf("%[^*]",line);
}
(a)Accepts the string that contains DIGITS & ALPHABETS
only.
(b)Accepts the string that contains * or asterisk characters
only.
(c)Accepts the string that contains anything other than the
* or asterisk character.
(d)None of these
Ans. (c)
_________________________________________________________________________________________________
[Q001]. What will be the output of the following program :
void main()
{
printf();
}
(a)Run-Time Error (b)Compile-Time Error (c)No Output (d)None of these
Ans. (b) Since there must be enough arguments for the
format.
_____________________________________________________
[Q002]. What will be the output of the following program :
void main()
{
printf(NULL);
}
(a)Run-Time Error (b)Compile-Time
Error (c)No Output (d)None of these
Ans. (c) Since NULL is a constant value or NULL pointer
value or a NULL string.
____________________________________________________
[Q003]. What will be the output of the following program :
void main()
{
printf("%%",7);
}
(a)7 (b)Compile-Time
Error (c)% (d)%%
Ans. (c) Since % is a format specifier & excess
arguments (more than required by the format) are
merely ignored.
__________________________________________________________
[Q004]. What will be the output of the following program :
void main()
{
printf("//",5);
}
(a)5 (b)Compile-Time
Error (c)/ (d)//
Ans. (d) Since // is taken as string
_____________________________________________________________
[Q005]. What will be the output of the following program :
void main()
{
printf("d%",8);
}
(a)8 (b)Compile-Time
Error (c)d% (d)None
of these
Ans. (c) Since excess arguments (more than required by the
format) are merely ignored.
__________________________________________________________
[Q006]. What will be the output of the following program :
void main()
{
printf("%d"+0,123);
}
(a)123 (b)Compile-Time Error (c)No Output (d)None of these
Ans. (a)
since"%d"+0 has no effect on the output operation.
__________________________________________________________________
[Q007]. What will be the output of the following program :
void main()-----------------doubt
{
printf("%d"+1,123);
}
(a)123 (b)Compile-Time
Error (c)d (d)No
Output
Ans. (c) since
"%d"+1 (i.e 1 or > 0) affects the program output by considering
"%d" as string and ignores 123
Where 1 refers to the index i.e. 2nd character in the array
or string "%d".
_______________________________________________________________
[Q008]. What will be the output of the following program :
void main()
{
printf("%d",printf("Hi!")+printf("Bye"));
}
(a)ByeHi!6 (b)Hi!Bye6 (c)Compile-Time
Error (d)None of these
Ans. (b) Since L->R priority & the length of the
strings 'Hi!' & 'Bye' is 3+3=6
____________________________________________________________
[Q009]. What will be the output of the following program :
void main()
{
printf("%d",printf("Hi!")*printf("Bye"));
}
(a)ByeHi!6 (b)Hi!Bye9 (c)Hi!Bye (d)None of these
Ans. (b) Since L->R priority & the length of the
strings 'Hi!' & 'Bye' is 3*3=9
________________________________________________________
[Q010]. What will be the output of the following program :
void main()
{
printf("%d",printf("")+printf(""));
}
(a)0 (b)No
Output (c)Compile-Time Error (d)None of these
Ans. (a) Since L->R priority & the length of the 2
empty strings are : 0+0=0
____________________________________________________________
[Q011]. What will be the output of the following program :
void main()
{
printf("Hi Friends"+3);
}
(a)Hi Friends (b)Friends (c)Hi
Friends3 (d)None of these
Ans. (b) Since (base adress)+0 points to the value 'H'. Now
the NEW (base address) equals (base address)+3 that points to the character
'F'. Thus it prints the string from 'F' onwards.
___________________________________________________________
[Q012]. What will be the output of the following program :
void main()
{
printf("C For ") + printf("Swimmers");
}
(a)Compile-Time Error (b)C For Swimmers (c)Run-Time
Error (d)None of these
Ans. (b) It is a VALID C statement.
____________________________________________________________
[Q013]. What will be the output of the following program :
void main()
{
printf("\/\*\-*\/");
}
(a)Run-Time Error (b)\/*-*\/ (c)/*-*/ (d)None of these
Ans. (c) Since \ is an escape sequence character. Be careful
while analyzing such statements.
_______________________________________________________________
[Q014]. What will be the output of the following program :
int main()
{
int main=7;
{
printf("%d",main);
return main;
}
printf("Bye");
}
(a)Compile-Time Error (b)Run-Time
Error (c)7Bye (d)7
Ans. (d) It is a VALID C statement. Prints 7 and returns the
same to the OS. NOTE: Last printf
statement will not be executed.
________________________________________________________
[Q015]. What will be the output of the following program :
void main()
{
main();
}
(a)Compile-Time Error (b)Run-Time
Error (c)Infinite
Loop (d)None of these
Ans. (c) It is a VALID C statement. It is like a recursive
function & the statements get executed infinite number of times. (All
compilers will not support)
_____________________________________________________________
[Q016]. What will be the output of the following program :
void main()
{
printf("Work" "Hard");
}
(a)Work (b)Hard (c)No Output (d)WorkHard
Ans. (d) Since L->R priority. First it prints the word
'Work' & then 'Hard'.
______________________________________________________________
[Q017]. What will be the output of the following program :
void main()
{
char
str[]="%d";
int val=25;
printf(str,val);
}
(a)Compile-Time Error (b)Run-Time
Error (c)25 (d)None of these
Ans. (c) It is a VALID C statement. First parameter contains
the format specifier & the Second
parameter contains the actual value 25.
________________________________________________________
[Q018]. What will be the output of the following program :
void main()
{
int val=75;
printf("%d",val,.,.);
}
(a)Compile-Time Error (b)Unpredictable (c)75 (d)None of these
Ans. (b) Output is Unpredictable B'coz there are not enough
arguments for the format. But it is a VALID C statement.
____________________________________________________
[Q019]. What will be the output of the following program :
void main()---------------------doubt
{
int val=10;
printf("%d",val+1,"%d",val--);
}
(a)10 (b)11
10 (c)11 9 (d)10 9
Ans. (a) Since
R->L priority. The second format specifier “%d” is an excess argument and it
is ignored.
______________________________________________________________
[Q020]. What will be the output of the following program :
void main()
{
int val=5;
printf("%d %d %d
%d",val,--val,++val,val--);
}
(a)3 4 6 5 (b)5
5 6 5 (c)4 4 5 5 (d)None
of these
Ans. (c) Since R->L priority.
___________________________________________________________
[Q021]. What will be the output of the following program :
void main()
{
int
val=5,num;
printf("%d",scanf("%d
%d",&val,&num));
}
[NOTE : ASSUME 2 values are entered by the user are stored
in the variables 'val' & 'num' respectively.]
(a)1 (b)2 (c)5 (d)None
of these
Ans. (b) Since scanf statement returns the number of input
fields successfully scanned, converted
& stored.
_________________________________________________________________________________________________
[Q022]. What will be the output of the following program :
#define
Compute(x,y,z) (x+y-z)---------------doubt
void main()
{
int
x=2,y=3,z=4;
printf("%d",Compute(y,z,(-x+y)) *
Compute(z,x,(-y+z)));
}
(a)40 (b)30 (c)Compile-Time
Error (d)None of
these
Ans. (b) Since it is macro function. NOTE : Be careful while
doing such type of calculations.
_________________________________________________________________________________________________
[Q023]. What will be the output of the following program :
void main()
{
int
m=10,n=20;
printf("%d %d %d",m/* m-value */,/*
n-value */n,m*/* Compute m*n */n);
}
(a)Run-Time Error (b)10 20 200 (c)Compile-Time
Error (d)None of
these
Ans. (b) Since comments /*...*/ are ignored by the compiler.
_________________________________________________________________________________________________
[Q024]. What will be the output of the following program :
void main()
{
int
m=10,n=20;
/* printf("%d",m*n);
}
(a)VALID but No Output (b)VALID : Prints 200 (c)Compile-Time
Error (d)None of
these
Ans. (c) Since COMMENT statement not ended properly i.e */
is missing in the above program.
_________________________________________________________________________________________________
[Q025]. What will be the output of the following program :
void main()
{
int val=97;
"Printing..."+printf("%c",val);
}
(a)Printing...97 (b)97 (c)Compile-Time
Error (d)a
Ans. (d) Since alphabet 'a' is the ASCII equivalent of 97.
_________________________________________________________________________________________________
[Q026]. What will be the output of the following program :
void main()
{
int val=5;
val=printf("C") +
printf("Skills");
printf("%d",val);
}
(a)Skills5 (b)C1 (c)Compile-Time Error (d)CSkills7
Ans. (d) VALID Since 'printf' function return the no. of
bytes output.
_________________________________________________________________________________________________
[Q027]. What will be the output of the following program :
void main()
{
char
str[]="Test";
if ((printf("%s",str)) == 4)
printf("Success");
else
printf("Failure");
}
(a)TestFailure (b)TestSuccess (c)Compile-Time
Error (d)Test
Ans. (b) VALID Since 'printf' function return the no. of
bytes output.
_________________________________________________________________________________________________
[Q028]. What will be the output of the following program :
void main()
{
int val=5;
printf("%*d",val);
}
(a) 5 (b)5 (c)Compile-Time
Error (d)None of these
Ans. (a) VALID Since '*' specifies the precision (i.e. the
next argument in the precision). If no
precision is specified then the value itself will be the
precision value. Thus it prints 5 BLANK
SPACES & then the value 5.
_________________________________________________________________________________________________
[Q029]. What will be the output of the following program :
void main()--------------------------doubt
{
int val=5;
printf("%d5",val);
}
(a)Compile-Time Error (b)5 (c)55 (d) 5
Ans. (c)
_________________________________________________________________________________________________
[Q030]. What will be the output of the following program :
void main()
}
int val=5;
printf("%d",5+val++);
{
(a)Compile-Time Error (b)5 (c)10 (d)11
Ans. (a) Since incorrect usage of pair of braces } and {.
Correct usage : Each compound statement
should be enclosed within a pair of braces, i.e { and }.
_________________________________________________________________________________________________
Topic
: Decision-making, Branching, Looping & Bit-wise operations
[Q001]. What will be
the output of the following program :
void main()
{
printf("Hi!");
if (-1)
printf("Bye");
}
(a)No Output (b)Hi! (c)Bye (d)Hi!Bye
Ans. (d)
_________________________________________________________________________________________________
[Q002]. What will be the output of the following program :
void main()
{
printf("Hi!");
if (0 || -1)
printf("Bye");
}
(a)No Output (b)Hi! (c)Bye (d)Hi!Bye
Ans. (d)
_________________________________________________________________________________________________
[Q003]. What will be the output of the following program :
void main()
{
printf("Hi!");
if (!1)
printf("Bye");
}
(a)Compile-Time error (b)Hi! (c)Bye (d)Hi!Bye
Ans. (b)
_________________________________________________________________________________________________
[Q004]. What will be the output of the following program :
void main()
{
printf("Hi!");
if !(0)
printf("Bye");
}
(a)Compile-Time error (b)Hi! (c)Bye (d)Hi!Bye
Ans. (a)
_________________________________________________________________________________________________
[Q005]. What will be the output of the following program :
void main()
{
printf("Hi!");
if
(-1+1+1+1-1-1-1+(-1)-(-1))
printf("Bye");
}
(a)No Output (b)Hi! (c)Bye (d)Hi!Bye
Ans. (d)
_________________________________________________________________________________________________
[Q006]. What will be the output of the following program :
void main()
{
if (sizeof(int) && sizeof(float)
&& sizeof(float)/2-sizeof(int))
printf("Testing");
printf("OK");
}
(a)No Output (b)OK (c)Testing (d)TestingOK
Ans. (b)
_________________________________________________________________________________________________
[Q007]. What will be the output of the following program :
void main()
{
int a=1,b=2,c=3,d=4,e;
if (e=(a & b | c ^ d))
printf("%d",e);
}
(a)0 (b)7 (c)3 (d)No Output
Ans. (b)
_________________________________________________________________________________________________
[Q008]. What will be the output of the following program :
void main()
{
unsigned val=0xffff;
if (~val)
printf("%d",val);
printf("%d",~val);
}
(a)Compile-Time error (b)-1 (c)0 (d)-1 0
Ans. (c)
_________________________________________________________________________________________________
[Q009]. What will be the output of the following program :
void main()
{
unsigned a=0xe75f,b=0x0EF4,c;
c=(a|b);
if ((c > a) && (c > b))
printf("%x",c);
}
(a)No Output (b)0xe75f (c)0xefff .(d)None of these
Ans. (c)
_________________________________________________________________________________________________
[Q010]. What will be the output of the following program :
void main()
{
unsigned val=0xabcd;
if (val>>16 | val<<16)
{
printf("Success");
return;
}
printf("Failure");
}
(a)No Output (b)Success (c)Failure (d)SuccessFailure
Ans.(b)
_________________________________________________________________________________________________
[Q011]. What will be the output of the following program :
void main()
{
unsigned x=0xf880,y=5,z;
z=x<<y;
printf("%#x %#x",z,x>>y-1);
}
(a)1000 f87 (b)8800
0xf88 (c)1000 f88 (d)0x1000 0xf88
Ans. (d)
_________________________________________________________________________________________________
[Q012]. What will be the output of the following program :
void main()
{
register int a=5;
int *b=&a;
printf("%d %d",a,*b);
}
(a)Compile-Time error (b)Run-Time
error (c)5 5 (d)Unpredictable
Ans. (a)
_________________________________________________________________________________________________
[Q013]. What will be the output of the following program :
auto int
a=5;
void main()
{
printf("%d",a);
}
(a)Compile-Time error (b)Run-Time
error (c)5 (d)Unpredictable
Ans. (a)
_________________________________________________________________________________________________
[Q014]. What will be the output of the following program :
void main()
{
auto int a=5;
printf("%d",a);
}
(a)Compile-Time error (b)Run-Time
error (c)5 (d)Unpredictable
Ans. (c)
_________________________________________________________________________________________________
[Q015]. What will be the output of the following program :
void main()
{
int a=1,b=2,c=3,d=4;
if (d > c)
if (c > b)
printf("%d
%d",d,c);
else if (c > a)
printf("%d
%d",c,d);
if (c > a)
if
(b < a)
printf("%d %d",c,a);
else
if (b < c)
printf("%d %d",b,c);
}
(a)4 3 3 4 (b)4
3 3 2 (c)4
32 3 (d)4 33 1
Ans. (c)
_________________________________________________________________________________________________
[Q016]. What will be the output of the following program :
void main()
{
int a=1,b=2,c=3,d=4;
if (d > c)
if (c > b)
printf("%d
%d",d,c);
if (c > a)
printf("%d
%d",c,d);
if (c > a)
if
(b < a)
printf("%d %d",c,a);
if
(b < c)
printf("%d %d",b,c);
}
(a)4 32 3 (b)4
33 42 3 (c)4 3
3 4 2 3 (d)None of these
Ans. (b)
_________________________________________________________________________________________________
[Q017]. What will be the output of the following program :
void main()
{
int a=1;
if (a == 2);
printf("C Program");
}
(a)No Output (b)C
Program (c)Compile-Time
Error
Ans. (b)
_________________________________________________________________________________________________
[Q018]. What will be the output of the following program :
void main()
{
int a=1;
if (a)
printf("Test");
else;
printf("Again");
}
(a)Again (b)Test (c)Compile-Time
Error (d)TestAgain
Ans. (d)
_________________________________________________________________________________________________
[Q019]. What will be the output of the following program :
void main()
{
int i=1;
for (; i<4; i++);
printf("%d\n",i);
}
(a)No Output (b)1 (c)4 (d)None of these
2
3
Ans. (c)
_________________________________________________________________________________________________
[Q020]. What will be the output of the following program :
void main()
{
int a,b;
for (a=0; a<10; a++);
for (b=25; b>9; b-=3);
printf("%d %d",a,b);
}
(a)Compile-Time error (b)10
9 (c)10
7 (d)None of these
Ans. (c)
_________________________________________________________________________________________________
[Q021]. What will be the output of the following program :
void main()
{
float i;
for (i=0.1; i<0.4; i+=0.1)
printf("%.1f",i);
}
(a)0.10.20.3 (b)Compile-Time
Error (c)Run-Time
Error (d)No Output
Ans. (a)
_________________________________________________________________________________________________
[Q022]. What will be the output of the following program :
void main()
{
int i;
for (i=-10; !i; i++);
printf("%d",-i);
}
(a)0 (b)Compile-Time
Error (c)10 (d)No Output
Ans. (c)
_________________________________________________________________________________________________
[Q023]. What will be the output of the following program :
void main()
{
int i=5;
do;
printf("%d",i--);
while (i>0);
}
(a)5 (b)54321 (c)Compile-Time
Error (d)None of these
Ans. (c)
_________________________________________________________________________________________________
[Q024]. What will be the output of the following program :
void main()
{
int i;
for (i=2,i+=2; i<=9; i+=2)
printf("%d",i);
}
(a)Compile-Time error (b)2468 (c)468 (d)None of these
Ans. (c)
_________________________________________________________________________________________________
[Q025]. What will be the output of the following program :
void main()
{
int i=3;
for (i--; i<7; i=7)
printf("%d",i++);
}
(a)No Output (b)3456 (c)23456 (d)None of these
Ans. (d)
_________________________________________________________________________________________________
[Q026]. What will be the output of the following program :
void main()
{
int i;
for (i=5; --i;)
printf("%d",i);
}
(a)No Output (b)54321 (c)4321 (d)None of
these
Ans. (c)
_________________________________________________________________________________________________
[Q027]. What will be the output of the following program :
void main()
{
int choice=3;
switch(choice)
{
default:
printf("Default");
case 1:
printf("Choice1");
break;
case 2:
printf("Choice2");
break;
}
}
(a)No Output (b)Default (c)DefaultChoice1 (d)None of these
Ans. (c)
_________________________________________________________________________________________________
[Q028]. What will be the output of the following program :
void main()
{
static int choice;
switch(--choice,choice-1,choice-1,choice+=2)
{
case 1:
printf("Choice1");
break;
case 2:
printf("Choice2");
break;
default:
printf("Default");
}
}
(a)Choice1 (b)Choice2 (c)Default (d)None of these
Ans. (a)
_________________________________________________________________________________________________
[Q029]. What will be the output of the following program :
void main()
{
for (;printf(""););
}
(a)Compile-Time error (b)Executes
ONLY once (c)Executes
INFINITELY (d)None of these
Ans. (b)
_________________________________________________________________________________________________
[Q030]. What will be the output of the following program :
void main()
{
int i;
for (;(i=4)?(i-4):i++;)
printf("%d",i);
}
(a)Compile-Time error (b)4 (c)Infinite
Loop (d)No Output
Ans. (d)
_________________________________________________________________________________________________
[Q031]. What will be the output of the following program :
void main()
{
static int j;
for (j<5; j<5; j+=j<5)
printf("%d",j++);
}
(a)024 (b)Compile-Time
Error (c)01234 (d)No Output
Ans.
_________________________________________________________________________________________________
[Q032]. What will be the output of the following program :
void main()
{
int i=9;
for (i--; i--; i--)
printf("%d ",i);
}
(a)9 6 3 (b)Compile-Time
Error (c)7 5 3 1 (d)Infinite Loop
Ans.
_________________________________________________________________________________________________
[Q033]. What will be the output of the following program :
void main()
{
int i;
for (i=5; ++i; i-=3)
printf("%d ",i);
}
(a)6 4 2 (b)Compile-Time
Error (c)6 3 1 (d)Infinite Loop
Ans.
_________________________________________________________________________________________________
[Q034]. Which of the following code causes INFINITE Loop :
(a)do while(1); (b)do;while(1); (c)do; (d)do{}while(1);
while(1);
(i)Only (a) (ii)Only
(b), (c) & (d) (iii)None
of these (iv)All of these
Ans.
_________________________________________________________________________________________________
[Q035]. What will be the output of the following program :
#define
Loop(i) for (j=0; j<i; j++){ \
sum += i+j; \
}
void main()
{
int i,j,sum=0;
for (i=0; i<=3; i++)
Loop(i)
printf("%d",sum);
}
(a)Run-Time Error (b)Compile-Time
Error (c)18 (d)0
Ans.
_________________________________________________________________________________________________
(1) What will be output if you will compile and execute the
following c code?
void main(){
int i=320;
char *ptr=(char *)&i;
printf("%d",*ptr);
}
(a)320
(b)1
(c)64
(d)Compiler error
(e)None of above
Output: (c)
Explanation:
As we know size of int data type is two byte while char
pointer can pointer one
byte at time.
Memory representation of int i=320
So char pointer ptr is pointing to only first byte as shown
above figure.
*ptr i.e. content of first byte is 01000000 and its decimal
value is 64.
How to represent char, int and float data in memory?
Data type tutorial.
(2) What will be output if you will compile and execute the
following c code?
#define x 5+2
void main(){
int i;
i=x*x*x;
printf("%d",i);
}
(a)343
(b)27
(c)133
(d)Compiler error
(e)None of above
Output: (b)
Explanation:
As we know #define is token pasting preprocessor it only
paste the value of
micro constant in the program before the actual compilation
start. If you will
see intermediate file you will find:
test.c 1:
test.c 2: void main(){
test.c 3: int i;
test.c 4: i=5+2*5+2*5+2;
test.c 5: printf("%d",i);
test.c 6: }
test.c 7:
You can absorb #define only pastes the 5+2 in place of x in
program. So,
i=5+2*5+2*5+2
=5+10+10+2
=27
What is intermediate file and how to see intermediate file?
Preprocessor tutorial.
(3) What will be output if you will compile and execute the
following c code?
void main(){
char c=125;
c=c+10;
printf("%d",c);
}
(a)135
(b)+INF
(c)-121
(d)-8
(e)Compiler error
Output: (c)
Explanation:
As we know char data type shows cyclic properties i.e. if
you will increase or
decrease the char variables beyond its maximum or minimum
value respectively it
will repeat same value according to following cyclic order:
So,
125+1= 126
125+2= 127
125+3=-128
125+4=-127
125+5=-126
125+6=-125
125+7=-124
125+8=-123
125+9=-122
125+10=-121
What is cyclic nature of data type?
Data type tutorial.
(4) What will be output if you will compile and execute the
following c code?
void main(){
float a=5.2;
if(a==5.2)
printf("Equal");
else if(a<5.2)
printf("Less than");
else
printf("Greater than");
}
(a)Equal
(b)Less than
(c)Greater than
(d)Compiler error
(e)None of above
Output: (b)
Explanation:
5.2 is double constant in c. In c size of double data is 8
byte while a is float
variable. Size of float variable is 4 byte.
So double constant 5.2 is stored in memory as:
101.00 11001100 11001100 11001100 11001100 11001100 11001101
Content of variable a will store in the memory as:
101.00110 01100110 01100110
It is clear variable a is less than double constant 5.2
Since 5.2 is recurring float number so it different for
float and double. Number
likes 4.5, 3.25, 5.0 will store same values in float and
double data type.
Note: In memory float and double data is stored in completely
different way. If
you want to see actual memory representation goes to
question number (60) and
(61).
Data type tutorial.
(5) What will be output if you will compile and execute the
following c code?
void main(){
int i=4,x;
x=++i + ++i + ++i;
printf("%d",x);
}
(a)21
(b)18
(c)12
(d)Compiler error
(e)None of above
Output: (a)
Explanation:
In ++a, ++ is pre increment operator. In any mathematical
expression pre
increment operator first increment the variable up to break
point then starts
assigning the final value to all variable.
Step 1: Increment the variable I up to break point.
Step 2: Start assigning final value 7 to all variable i in
the expression.
So, i=7+7+7=21
What is break point?
Operator tutorial.
(6) What will be output if you will compile and execute the
following c code?
void main(){
int a=2;
if(a==2){
a=~a+2<<1;
printf("%d",a);
}
else
{ break;
}
}
(a)It will print nothing.
(b)-3
(c)-2
(d)1
(e)Compiler error
Output: (e)
Explanation:
Keyword break is not part of if-else statement. Hence it
will show compiler
error: Misplaced break
Where we can use break keyword?
Control statement tutorial
(7) What will be output if you will compile and execute the
following c code?
void main(){
int a=10;
printf("%d %d %d",a,a++,++a);
}
(a)12 11 11
(b)12 10 10
(c)11 11 12
(d)10 10 12
(e)Compiler error
Output: (a)
Explanation:
In c printf function follows cdecl parameter passing scheme.
In this scheme
parameter is passed from right to left direction.
So first ++a will pass and value of variable will be a=10
then a++ will pass now
value variable will be a=10 and at the end a will pass and
value of a will be
a=12.
What is cedecl and pascal parameter passing convention?
Function tutorial.
(8) What will be output if you will compile and execute the
following c code?
void main(){
char *str="Hello world";
printf("%d",printf("%s",str));
}
(a) 11Hello world
(b) 10Hello world
(c) Hello world10
(d) Hello world11
(e) Compiler error
Output: (d)
Explanation:
Return type of printf function is integer and value of this
integer is exactly
equal to number of character including white space printf
function prints. So,
printf(“Hello world”) will return 13.
What is prototype of printf function?
Formatted I/O tutorial.
(9) What will be output if you will compile and execute the
following c code?
#include "stdio.h"
#include "string.h"
void main(){
char *str=NULL;
strcpy(str,"cquestionbank");
printf("%s",str);
}
(a)cquestionbank
(b)cquestionbank\0
(c)(null)
(d)It will print nothing
(e)Compiler error
Output: (c)
Explanation:
We cannot copy any thing using strcpy function to the
character pointer pointing
to NULL.
String tutorial.
More questions of string.
(10) What will be output if you will compile and execute the
following c code?
#include "stdio.h"
#include "string.h"
void main(){
int i=0;
for(;i<=2;)
printf(" %d",++i);
}
(a)0 1 2
(b)0 1 2 3
(c)1 2 3
(d)Compiler error
(e)Infinite loop
Output: (c)
Explanation:
In for loop each part is optional.
Complete tutorial of looping in C.
(11) What will be output if you will compile and execute the
following c code?
void main(){
int x;
for(x=1;x<=5;x++);
printf("%d",x);
}
(a)4
(b)5
(c)6
(d)Compiler error
(e)None of above
Output: (c)
Explanation:
Body of for loop is optional. In this question for loop will
execute until value
of variable x became six and condition became false.
Looping tutorial.
(12) What will be output if you will compile and execute the
following c code?
void main(){
printf("%d",sizeof(5.2));
}
(a)2
(b)4
(c)8
(d)10
(e)Compiler error
Output: (c)
Explanation:
Default type of floating point constant is double. So 5.2 is
double constant and
its size is 8 byte.
Detail explanation of all types of constant in C.
(13) What will be output if you will compile and execute the
following c code?
#include "stdio.h"
#include "string.h"
void main(){
char c='\08';
printf("%d",c);
}
(a)8
(b)’8’
(c)9
(d)null
(e)Compiler error
Output: (e)
Explanation:
In c any character is starting with character ‘\’ represents
octal number in
character. As we know octal digits are: 0, 1, 2, 3, 4, 5, 6,
and 7. So 8 is not
an octal digit. Hence ‘\08’ is invalid octal character
constant.
Octal character constantan.
Hexadecimal character constant.
(14) What will be output if you will compile and execute the
following c code?
#define call(x,y) x##y
void main(){
int x=5,y=10,xy=20;
printf("%d",xy+call(x,y));
}
(a)35
(b)510
(c)15
(d)40
(e)None of above
Output: (d)
Explanation:
## is concatenation c preprocessor operator. It only
concatenates the operands
i.e.
a##b=ab
If you will see intermediate file then you will find code
has converted into
following intermediate code before the start of actual
compilation.
Intermediate file:
test.c 1:
test.c 2: void main(){
test.c 3: int x=5,y=10,xy=20;
test.c 4: printf("%d",xy+xy);
test.c 5: }
test.c 6:
It is clear call(x, y) has replaced by xy.
What is macro call?
Preprocessor tutorial.
(15) What will be output if you will compile and execute the
following c code?
int * call();
void main(){
int *ptr;
ptr=call();
clrscr();
printf("%d",*ptr);
}
int * call(){
int a=25;
a++;
return &a;
}
(a)25
(b)26
(c)Any address
(d)Garbage value
(e)Compiler error
Output: (d)
Explanation:
In this question variable a is a local variable and its
scope and visibility is
within the function call. After returning the address of a
by function call
variable a became dead while pointer ptr is still pointing
to address of
variable a. This problem is known as dangling pointer
problem.
Complete pointer tutorial.
(16) What is error in following declaration?
struct outer{
int a;
struct inner{
char c;
};
};
(a)Nesting of structure is not allowed in c.
(b)It is necessary to initialize the member variable.
(c)Inner structure must have name.
(d)Outer structure must have name.
(e)There is not any error.
Output: (c)
Explanation:
It is necessary to assign name of inner structure at the
time of declaration
other wise we cannot access the member of inner structure.
So correct
declaration is:
struct outer{
int a;
struct inner{
char c;
}name;
};
Structure tutorial.
Union tutorial.
(17) What will be output if you will compile and execute the
following c code?
void main(){
int array[]={10,20,30,40};
printf("%d",-2[array]);
}
(a)-60
(b)-30
(c)60
(d)Garbage value
(e)Compiler error
Output: (b)
Explanation:
In c,
array[2]=*(array+2)=*(2+array)=2[array]=30
Array tutorial.
Array of pointer.
How to read complex pointers.
(18) What will be output if you will compile and execute the
following c code?
void main(){
int i=10;
static int x=i;
if(x==i)
printf("Equal");
else if(x>i)
printf("Greater than");
else
printf("Less than");
}
(a)Equal
(b)Greater than
(c)Less than
(d)Compiler error
(e)None of above
Output: (d)
Explanation:
static variables are load time entity while auto variables
are run time entity.
We can not initialize any load time variable by the run time
variable.
In this example i is run time variable while x is load time
variable.
What is storage class?
(18) What will be output if you will compile and execute the
following c code?
void main(){
int i=5,j=2;
if(++i>j++||i++>j++)
printf("%d",i+j);
}
(a)7
(b)11
(c)8
(d)9
(e)Compiler error
Output: (d)
Explanation:
|| is logical OR operator. In C logical OR operator doesn’t
check second operand
if first operand is true.
++i>j++ || i++>j++
First operand: ++i>j++
Second operand: i++>j++
First operand
++i > j++
=> 6 > 2
Since first operand is true so it will not check second
operand.
Hence i= 6 and j=3
Properties of && operator.
Operator tutorial with examples.
(19) What will be output if you will compile and execute the
following c code?
#define max 5;
void main(){
int i=0;
i=max++;
printf("%d",i++);
}
(a)5
(b)6
(c)7
(d)0
(e)Compiler error
Output: (e)
Explanation:
#define is token pasting preprocessor. If you will see
intermediate file: test.i
test.c 1:
test.c 2: void main(){
test.c 3: int i=0;
test.c 4: i=5++;
test.c 5: printf("%d",i++);
test.c 6: }
test.c 7:
It is clear macro constant max has replaced by 5. It is
illegal to increment the
constant number. Hence compiler will show Lvalue required.
What is Lvalue and Rvalue?
How to see intermediate file?
Preprocessor questions and answer.
(20) What will be output if you will compile and execute the
following c code?
void main(){
double far* p,q;
printf("%d",sizeof(p)+sizeof q);
}
(a)12
(b)8
(c)4
(d)1
(e)Compiler error
Output: (a)
Explanation:
It is clear p is far pointer and size of far pointer is 4
byte while q is double
variable and size of double variable is 8 byte.
What is near pointer?
What is far pointer?
What is huge pointer?
Complete pointer tutorial.
(21) What will be output if you will compile and execute the
following c code?
void main(){
int a=5;
float b;
printf("%d",sizeof(++a+b));
printf(" %d",a);
}
(a)2 6
(b)4 6
(c)2 5
(d)4 5
(e)Compiler error
Output: (d)
Explanation:
++a +b
=6 + Garbage floating point number
=Garbage floating point number
//From the rule of automatic type conversion
Hence sizeof operator will return 4 because size of float
data type in c is 4
byte.
Value of any variable doesn’t modify inside sizeof operator.
Hence value of
variable a will remain 5.
Properties of sizeof operator.
Operators tutorial
(22) What will be output if you will compile and execute the
following c code?
void main(){
char huge *p=(char *)0XC0563331;
char huge *q=(char *)0XC2551341;
if(p==q)
printf("Equal");
else if(p>q)
printf("Greater than");
else
printf("Less than");
}
(a)Equal
(b)Greater than
(c)Less than
(d)Compiler error
(e)None of above
Output: (a)
Explanation:
As we know huge pointers compare its physical address.
Physical address of huge pointer p
Huge address: 0XC0563331
Offset address: 0x3331
Segment address: 0XC056
Physical address= Segment address * 0X10 + Offset address
=0XC056 * 0X10 +0X3331
=0XC0560 + 0X3331
=0XC3891
Physical address of huge pointer q
Huge address: 0XC2551341
Offset address: 0x1341
Segment address: 0XC255
Physical address= Segment address * 0X10 + Offset address
=0XC255 * 0X10 +0X1341
=0XC2550 + 0X1341
=0XC3891
Since both huge pointers p and q are pointing same physical
address so if
condition will true.
Great article.
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