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Home » Education » 200 Best C C++ FAQs For IIT, JEE, OJEE and Freshers for Campus Interview with Explanation

200 Best C C++ FAQs For IIT, JEE, OJEE and Freshers for Campus Interview with Explanation


200 Best C C++ FAQs For IIT, JEE, OJEE and Freshers for Campus Interview with Explanation

 Note : All the programs are tested under Turbo C/C++ compilers.

It is assumed that,

  • Programs run under DOS environment,
  • The underlying machine is an x86 system,
  • rogram 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).

Predict the output or error(s) for the following:

 

1.      void main()

{

            int  const * p=5;

            printf(“%d”,++(*p));

}

Answer:

Compiler error: Cannot modify a constant value.

Explanation:

p

is a pointer to a “constant integer”. But we tried to change the
value of the “constant integer”.

2.      main()

{

            char
s[ ]=”man”;

            int
i;

            for(i=0;s[
i ];i++)

            printf(“\n%c%c%c%c”,s[
i ],*(s+i),*(i+s),i[s]);

}

Answer:  mmmm

aaaa

nnnn

Explanation:

s[i],
*(i+s), *(s+i), i[s] are all different ways of expressing the same idea.
Generally  array name is the base address
for that array. Here s is the base
address. i is the index number/displacement
from the base address. So, indirecting it with * is same as s[i]. i[s] may be
surprising. But in the  case of  C  it
is same as s[i].

3.      main()

{

            float
me = 1.1;

            double
you = 1.1;

            if(me==you)

printf(“I
love U”);

else

                        printf(“I
hate U”);

}

Answer:
I
hate U

Explanation:

For
floating point numbers (float,
double, long double) the values
cannot be predicted exactly. Depending on the number of bytes, the precession
with of the value  represented varies.
Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with
less precision than long double.

Rule of Thumb:

Never
compare or at-least be cautious when using floating point numbers with
relational operators (== , >,
<, <=, >=,!= ) .

4.      main()

            {

            static
int var = 5;

            printf(“%d
“,var–);

            if(var)

                        main();

            }

Answer:

5
4 3 2 1

Explanation:

When
static storage class is given, it is
initialized once. The change in the value of a static variable is retained even between the function calls. Main
is also treated like any other ordinary function, which can be called
recursively.

5.      main()

{

             int c[ ]={2.8,3.4,4,6.7,5};

             int j,*p=c,*q=c;

             for(j=0;j<5;j++) {

                        printf(”
%d “,*c);

                        ++q;     }

             for(j=0;j<5;j++){

printf(”
%d “,*p);

++p;     }

}

Answer:

2
2 2 2 2 2 3 4 6 5

Explanation:

Initially
pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value
2 will be printed 5 times. In second loop p
itself is incremented. So the values 2 3 4 6 5 will be printed.

6.      main()

{

            extern
int i;

            i=20;

printf(“%d”,i);

}

 

Answer:

Linker Error
: Undefined symbol ‘_i’

Explanation:

extern
storage class in the following declaration,

extern int i;

specifies
to the compiler that the memory for i
is allocated in some other program and that address will be given to the
current program at the time of linking. But linker finds that no other variable
of name i is available in any other
program with memory space allocated for it. Hence a linker error has occurred .

 

7.      main()

{

            int
i=-1,j=-1,k=0,l=2,m;

            m=i++&&j++&&k++||l++;

            printf(“%d
%d %d %d %d”,i,j,k,l,m);

}

Answer:

0
0 1 3 1

Explanation
:

Logical
operations always give a result of 1 or
0
. And also the logical AND (&&) operator has higher priority over
the logical OR (||) operator. So the expression
i++ && j++ &&
k++’
is executed first. The result of this expression is 0    (-1 && -1 && 0 = 0). Now
the expression is 0 || 2 which evaluates to 1 (because OR operator always gives
1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is
1. The values of other variables are also incremented by 1.

 

8.      main()

{

            char
*p;

            printf(“%d
%d “,sizeof(*p),sizeof(p));

}

 

Answer:

1
2

Explanation:

The
sizeof() operator gives the number of bytes taken by its operand. P is a
character pointer, which needs one byte for storing its value (a character).
Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the
address of the character pointer sizeof(p) gives 2.

 

9.      main()

{

            int
i=3;

            switch(i)

             {

                default:printf(“zero”);

                case 1: printf(“one”);

                           break;

               case 2:printf(“two”);

                          break;

              case 3: printf(“three”);

                          break;

              } 

}

Answer
:

three

Explanation
:

The
default case can be placed anywhere inside the loop. It is executed only when
all other cases doesn’t match.

 

10.  main()

{

              printf(“%x”,-1<<4);

}

Answer:

fff0

Explanation
:

-1
is internally represented as all 1’s. When left shifted four times the least
significant 4 bits are filled with 0’s.The %x format specifier specifies that
the integer value be printed as a hexadecimal value.

 

11.  main()

{


char string[]=”Hello World”;

            display(string);

}

void display(char *string)

{

printf(“%s”,string);

}

            Answer:

Compiler Error
:
Type mismatch in redeclaration of function display

Explanation
:

In
third line, when the function display
is encountered, the compiler doesn’t know anything about the function display.
It assumes the arguments and return types to be integers, (which is the default
type). When it sees the actual function display,
the arguments and type contradicts with what it has assumed previously. Hence a
compile time error occurs.

 

12.  main()

{

            int
c=- -2;

            printf(“c=%d”,c);

}

Answer:

c=2;

Explanation:

Here
unary minus (or negation) operator is used twice. Same maths  rules applies, ie. minus * minus= plus.

Note:

However
you cannot give like –2. Because — operator can  only be applied to variables as a decrement operator (eg., i–). 2 is a
constant and not a variable.

 

13.  #define int char

main()

{

            int
i=65;

            printf(“sizeof(i)=%d”,sizeof(i));

}

Answer:

sizeof(i)=1

Explanation:

Since
the #define replaces the string  int by the macro char

 

14.  main()

{

int i=10;

i=!i>14;

Printf (“i=%d”,i);

}

Answer:

i=0

Explanation:

In
the expression !i>14 , NOT (!)
operator has more precedence than ‘ >’ symbol.  !
is a unary logical operator. !i (!10) is 0 (not of true is false).  0>14 is false (zero).

 

15.  #include<stdio.h>

main()

{

char
s[]={‘a’,’b’,’c’,’\n’,’c’,’\0′};

char *p,*str,*str1;

p=&s[3];

str=p;

str1=s;

printf(“%d”,++*p +
++*str1-32);

}

Answer:

77

Explanation:

p
is pointing to character ‘\n’. str1 is pointing to character ‘a’ ++*p. “p
is pointing to ‘\n’ and that is incremented by one.” the ASCII value of
‘\n’ is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1,
str1 is pointing to ‘a’ that is incremented by 1 and it becomes ‘b’. ASCII
value of ‘b’ is 98.

Now performing (11 + 98 – 32), we get
77(“M”);

So we get the output 77 :: “M”
(Ascii is 77).

 

16.  #include<stdio.h>

main()

{

int a[2][2][2] = { {10,2,3,4},
{5,6,7,8}  };

int *p,*q;

p=&a[2][2][2];

*q=***a;

printf(“%d—-%d”,*p,*q);

}

Answer:

SomeGarbageValue—1

Explanation:

p=&a[2][2][2]  you declare only two 2D arrays, but you are
trying to access the third 2D(which you are not declared) it will print garbage
values. *q=***a starting address of a is assigned integer pointer. Now q is
pointing to starting address of a. If you print *q, it will print first element
of 3D array.

17.  #include<stdio.h>

main()

{

struct xx

{

      int x=3;

      char name[]=”hello”;

 };

struct xx *s;

printf(“%d”,s->x);

printf(“%s”,s->name);

}

Answer:

Compiler
Error

Explanation:

You
should not initialize variables in declaration

 

18.  #include<stdio.h>

main()

{

struct xx

{

int x;

struct yy

{

char s;

            struct
xx *p;

};

struct yy *q;

};

}

Answer:

Compiler
Error

Explanation:

The
structure yy is nested within structure xx. Hence, the elements are of yy are
to be accessed through the instance of structure xx, which needs an instance of
yy to be known. If the instance is created after defining the structure the
compiler will not know about the instance relative to xx. Hence for nested
structure yy you have to declare member.

 

19.  main()

{

printf(“\nab”);

printf(“\bsi”);

printf(“\rha”);

}

Answer:

hai

Explanation:

\n  – newline

\b  – backspace

\r  – linefeed

 

20.  main()

{

int i=5;

printf(“%d%d%d%d%d%d”,i++,i–,++i,–i,i);

}

Answer:

45545

Explanation:

The
arguments in a function call are pushed into the stack from left to right. The
evaluation is by popping out from the stack. and the  evaluation is from right to left, hence the
result.

 

21.  #define square(x) x*x

main()

{

int i;

i = 64/square(4);

printf(“%d”,i);

}

Answer:

64

Explanation:

the
macro call square(4) will substituted by 4*4 so the expression becomes i =
64/4*4 . Since / and * has equal priority the expression will be evaluated as
(64/4)*4 i.e. 16*4 = 64

 

22.  main()

{

char *p=”hai friends”,*p1;

p1=p;

while(*p!=’\0′) ++*p++;

printf(“%s   %s”,p,p1);

}

Answer:

ibj!gsjfoet

            Explanation:

                        ++*p++
will be parse in the given order

  • *p that is value at the location currently
    pointed by p will be taken
  • ++*p the retrieved value will be incremented
  • when ; is encountered the location will be
    incremented that is p++ will be executed

Hence, in the while loop initial
value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and
pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on.
Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes
“ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print
anything.

 

23.  #include <stdio.h>

#define a 10

main()

{

#define a 50

printf(“%d”,a);

}

Answer:

50

Explanation:

The
preprocessor directives can be redefined anywhere in the program. So the most
recently assigned value will be taken.

 

24.  #define clrscr() 100

main()

{

clrscr();

printf(“%d\n”,clrscr());

}

Answer:

100

Explanation:

Preprocessor
executes as a seperate pass before the execution of the compiler. So textual
replacement of clrscr() to 100 occurs.The input
program to compiler looks like this :

main()

{

100;

printf(“%d\n”,100);

}

Note:

100;
is an executable statement but with no action. So it doesn’t give any problem

 

25.  main()

{

printf(“%p”,main);

}

Answer:

Some address will be
printed.

Explanation:

Function
names are just addresses (just like array names are addresses).

main() is also
a function. So the address of function main will be printed. %p in printf
specifies that the argument is an address. They are printed as hexadecimal
numbers.

 

27)       main()

{

clrscr();

}

clrscr();

Answer:

No
output/error

Explanation:

The
first clrscr() occurs inside a function. So it becomes a function call. In the
second clrscr(); is a function declaration (because it is not inside any
function).

 

28)       enum colors {BLACK,BLUE,GREEN}

main()

{

 

printf(“%d..%d..%d”,BLACK,BLUE,GREEN);

return(1);

}

Answer:

0..1..2

Explanation:

enum
assigns numbers starting from 0, if not explicitly defined.

 

29)       void main()

{

char far *farther,*farthest;

 

printf(“%d..%d”,sizeof(farther),sizeof(farthest));

}

Answer:

4..2

Explanation:

the
second pointer is of char type and not a far pointer

 

30)       main()

{

int i=400,j=300;

printf(“%d..%d”);

}

Answer:

400..300

Explanation:

printf
takes the values of the first two assignments of the program. Any number of
printf’s may be given. All of them take only the first two values. If more
number of assignments given in the program,then printf will take garbage
values.

31)       main()

{

char *p;

p=”Hello”;

printf(“%c\n”,*&*p);

}

Answer:

H

Explanation:

*
is a dereference operator & is a reference
operator. They can be    applied
any number of times provided it is meaningful. Here  p points to
the first character in the string “Hello”. *p dereferences it
and so its value is H. Again  &
references it to an address and * dereferences it to the value H.

 

32)       main()

{

int i=1;

while (i<=5)

{

printf(“%d”,i);

if (i>2)
goto here;

i++;

}

}

fun()

{

here:

printf(“PP”);

}

Answer:

Compiler
error: Undefined label ‘here’ in function main

Explanation:

Labels
have functions scope, in other words The scope of the labels is limited to
functions . The label ‘here’ is available in function fun() Hence it is not
visible in function main.

 

33)       main()

{

static char
names[5][20]={“pascal”,”ada”,”cobol”,”fortran”,”perl”};

int i;

char *t;

t=names[3];

names[3]=names[4];

names[4]=t;

for (i=0;i<=4;i++)

printf(“%s”,names[i]);

}

Answer:

Compiler
error: Lvalue required in function main

Explanation:

Array
names are pointer constants. So it cannot be modified.

 

34)       void main()

{

int i=5;

printf(“%d”,i++ + ++i);

}

Answer:

Output
Cannot be predicted  exactly.

Explanation:

Side
effects are involved in the evaluation of
i

35)       void main()

{

int i=5;

printf(“%d”,i+++++i);

}

Answer:

Compiler
Error

Explanation:

The
expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of
operators.

36)       #include<stdio.h>

main()

{

int
i=1,j=2;

switch(i)

{

case 1:
printf(“GOOD”);
break;

case j:
printf(“BAD”);

break;

}

}

Answer:

Compiler
Error: Constant expression required in function main.

Explanation:

The
case statement can have only constant expressions (this implies that we cannot
use variable names directly so an error).

Note:

Enumerated
types can be used in case statements.

37)       main()

{

int
i;

printf(“%d”,scanf(“%d”,&i));  // value 10 is given as input here

}

Answer:

1

Explanation:

Scanf
returns number of items successfully read and not 1/0.  Here 10 is given as input which should have
been scanned successfully. So number of items read is 1.

38)       #define f(g,g2) g##g2

main()

{

int
var12=100;

printf(“%d”,f(var,12));

}

Answer:

100

39)       main()

{

int
i=0;

 

for(;i++;printf(“%d”,i))
;

printf(“%d”,i);

}

Answer:

            1

Explanation:

before
entering into the for loop the checking condition is “evaluated”.
Here it evaluates to 0 (false) and comes out of the loop, and i is incremented
(note the semicolon after the for loop).

40)       #include<stdio.h>

main()

{

char s[]={‘a’,’b’,’c’,’\n’,’c’,’\0′};

char *p,*str,*str1;

p=&s[3];

str=p;

str1=s;

printf(“%d”,++*p + ++*str1-32);

}

Answer:

M

Explanation:

p
is pointing to character ‘\n’.str1 is pointing to character ‘a’ ++*p meAnswer:”p
is pointing to ‘\n’ and that is incremented by one.” the ASCII value of
‘\n’ is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1
meAnswer:”str1 is pointing to ‘a’ that is incremented by 1 and it becomes
‘b’. ASCII value of ‘b’ is 98. both 11 and 98 is added and result is subtracted
from 32.

i.e.
(11+98-32)=77(“M”);

41)       #include<stdio.h>

main()

{

struct xx

{

int x=3;

char name[]=”hello”;

};

struct
xx *s=malloc(sizeof(struct xx));

printf(“%d”,s->x);

printf(“%s”,s->name);

}

Answer:

Compiler
Error

Explanation:

Initialization
should not be done for structure members inside the structure declaration

42)       #include<stdio.h>

main()

{

struct
xx

{
int x;
struct yy
{
char s;
struct xx *p;
};

struct yy *q;

};

}

Answer:

Compiler
Error

Explanation:

in
the end of nested structure yy a member have to be declared.

43)       main()

{

extern int i;

i=20;

printf(“%d”,sizeof(i));

}

Answer:

Linker
error: undefined symbol ‘_i’.

Explanation:

extern
declaration specifies that the variable i is defined somewhere else. The
compiler passes the external variable to be resolved by the linker. So compiler
doesn’t find an error. During linking the linker searches for the definition of
i. Since it is not found the linker flags an error.

44)       main()

{

printf(“%d”,
out);

}

int
out=100;

Answer:

Compiler
error: undefined symbol out in function main.

Explanation:

The
rule is that a variable is available for use from the point of declaration.
Even though a is a global variable, it is not available for main. Hence an
error.

45)       main()

{

extern out;

printf(“%d”, out);

}

int out=100;

Answer:

100

Explanation:

This
is the correct way of writing the previous program.

46)       main()

{

show();

}

void
show()

{

printf(“I’m the greatest”);

}

Answer:

Compier
error: Type mismatch in redeclaration of show.

Explanation:

When
the compiler sees the function show it doesn’t know anything about it. So the
default return type (ie, int) is assumed. But when compiler sees the actual
definition of show mismatch occurs since it is declared as void. Hence the
error.

The
solutions are as follows:

1.
declare void show() in main() .

2.
define show() before main().

3.
declare extern void show() before the use of show().

 

47)       main( )

{

int a[2][3][2] =
{{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};

printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);

}

Answer:

100,
100, 100, 2

114, 104, 102, 3

Explanation:

The given array is a 3-D one.
It can also be viewed as a 1-D array.

2 4 7 8 3 4 2 2 2 3 3 4

100
102  104  106 108
110  112  114
116   118   120
122

 

thus, for the
first printf statement a, *a, **a  give
address of  first element . since the
indirection ***a gives the value. Hence, the first line of the output.

for the second
printf a+1 increases in the third dimension thus points to value at 114, *a+1
increments in second dimension thus points to 104, **a +1 increments the first
dimension thus points to 102 and ***a+1 first gets the value at first location
and then increments it by 1. Hence, the output.

 

48)       main( )

{

int a[ ] = {10,20,30,40,50},j,*p;

for(j=0; j<5; j++)

{

printf(“%d”
,*a);

a++;

}

p = a;

for(j=0; j<5; j++)

{

printf(“%d
” ,*p);

p++;

}

}

Answer:

Compiler
error: lvalue required.

                       

Explanation:

Error is in
line with statement a++. The operand must be an lvalue and may be of any of
scalar type for the any operator, array name only when subscripted is an
lvalue. Simply array name is a non-modifiable lvalue.

 

49)       main( )

{

static int
a[ ]   = {0,1,2,3,4};

int  *p[
] = {a,a+1,a+2,a+3,a+4};

int
**ptr =  p;

ptr++;

printf(“\n %d
%d  %d”, ptr-p, *ptr-a, **ptr);

*ptr++;

printf(“\n %d
%d  %d”, ptr-p, *ptr-a, **ptr);

*++ptr;

printf(“\n %d  %d
%d”, ptr-p, *ptr-a, **ptr);

++*ptr;

printf(“\n %d
%d  %d”, ptr-p, *ptr-a, **ptr);

}

Answer:

            111

            222

            333

            344

Explanation:

Let us consider the
array and the two pointers with some address

a   

0 1 2 3 4

100      102
104      106      108

p

100 102 104 106 108

1000
1002    1004    1006
1008
ptr 

1000

2000

After execution of the instruction ptr++
value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr –
p is value in ptr – starting location of array p, (1002 – 1000) / (scaling
factor) = 1,  *ptr – a = value at address
pointed by ptr – starting value of array a, 1002 has a value 102  so the value is (102 – 100)/(scaling factor)
= 1,  **ptr is the value stored in the
location pointed by  the pointer of ptr =
value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the
output of the firs printf is  1, 1, 1.

After execution of *ptr++ increments value of
the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for
the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.

After execution of *++ptr increments value of
the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for
the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.

After execution of ++*ptr value in ptr
remains the same, the value pointed by the value is incremented by the scaling
factor. So the value in array p at location 1006 changes from 106 10 108,.
Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr –
a = 108 – 100 = 4, **ptr = 4.

 

50)       main( )

{

char
*q;

int  j;

for (j=0; j<3; j++) scanf(“%s” ,(q+j));

for (j=0; j<3; j++) printf(“%c” ,*(q+j));

for (j=0; j<3; j++) printf(“%s” ,(q+j));

}

Explanation:

Here we have
only one pointer to type char and since we take input in the same pointer thus
we keep writing over in the same location, each time shifting the pointer value
by 1. Suppose the inputs are MOUSE,
TRACK and VIRTUAL. Then for the first input suppose the pointer starts
at location 100 then the input one is stored as

M

O

U

S E \0

When the second
input is given the pointer is incremented as j value becomes 1, so the input is
filled in memory starting from 101.

M

T

R

A

C

K

\0

The
third input  starts filling from the
location 102

M

T

V

I

R

T

U

A

L

\0

This is the final value stored .

The
first printf prints the values at the position q, q+1 and q+2  = M T V

The
second printf prints three strings starting from locations q, q+1, q+2

i.e  MTVIRTUAL, TVIRTUAL and VIRTUAL.

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