C pointers questions



C pointers interview questions and answers

Frequently asked technical objective types multiple choice pointer questions with explanation of placement in  c programming language
Note: Linux GCC compilers and Visual C++ compiler doesn't support far and huge pointers.


1.
What will be output of following program?

#include<stdio.h>
int main(){
   int a = 320;
   char *ptr;
   ptr =( char *)&a;
   printf("%d ",*ptr);
   return 0;
}
(A) 2
(B) 320
(C) 64
(D) Compilation error
(E) None of above


Explanation:



Turbo C++ 3.0: 64

Turbo C ++4.5: 64

Linux GCC: 64

Visual C++: 64



As we know int is two byte data byte while char is one byte data byte. char pointer can keep the address one byte at time.

Binary value of 320 is 00000001 01000000 (In 16 bit)

Memory representation of int a = 320 is:




So ptr is pointing only first 8 bit which color is green and Decimal value is 64.

2.
What will be output of following program?

#include<stdio.h>
#include<conio.h>
int main(){
   void (*p)();
   int (*q)();
   int (*r)();
   p = clrscr;
   q = getch;
   r = puts;
  (*p)();
  (*r)("cquestionbank.blogspot.com");
  (*q)();
  return 0;
}

(A) NULL
(B) cquestionbank.blogspot.com
(C) c
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: cquestionbank.blogspot.com

Turbo C ++4.5: cquestionbank.blogspot.com

Linux GCC: Compilation error

Visual C++: Compilation error



p is pointer to function whose parameter is void and return type is also void. r and q is pointer to function whose parameter is void and return type is int . So they can hold the address of such function.

3.
What will be output of following program?

#include<stdio.h>
int main(){
   int i = 3;
   int *j;
   int **k;
   j=&i;
   k=&j;
   printf("%u %u %d ",k,*k,**k);
   return 0;
}
(A) Address, Address, 3
(B) Address, 3, 3
(C) 3, 3, 3
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: Address, Address, 3

Turbo C ++4.5: Address, Address, Address

Linux GCC: Address, Address, 3

Visual C++: Address, Address, 3



Memory representation




Here 6024, 8085, 9091 is any arbitrary address, it may be different.

Value of k is content of k in memory which is 8085

Value of *k means content of memory location which address k keeps.

k keeps address 8085 .

Content of at memory location 8085 is 6024

In the same way **k will equal to 3.

Short cut way to calculate:

Rule: * and & always cancel to each other

i.e. *&a = a

So *k = *(&j) since k = &j

*&j = j = 6024

And
**k = **(&j) = *(*&j) = *j = *(&i) = *&i = i = 3

4.
What will be output of following program?

#include<stdio.h>
int main(){
   char far *p =(char far *)0x55550005;
   char far *q =(char far *)0x53332225;
   *p = 80;
   (*p)++;
   printf("%d",*q);
   return 0;
}
(A) 80
(B) 81
(C) 82
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: 81

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error



Far address of p and q are representing same physical address.

Physical address of 0x55550005 = (0x5555) * (0x10) + (0x0005) = 0x55555

Physical address of 0x53332225 = (0x5333 * 0x10) + (0x2225) = 0x55555

*p = 80, means content at memory location 0x55555 is assigning value 25

(*p)++ means increase the content by one at memory location 0x5555 so now content at memory location 0x55555 is 81

*q also means content at memory location 0x55555 which is 26

5.
What will be output of following program?

#include<stdio.h>
#include<string.h>
int main(){
char *ptr1 = NULL;
char *ptr2 = 0;
strcpy(ptr1," c");
strcpy(ptr2,"questions");
printf("\n%s %s",ptr1,ptr2);
return 0;
}
(A) c questions
(B) c (null)
(C) (null) (null)
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: (null) (null)

Turbo C ++4.5: Run time error

Linux GCC: Run time error

Visual C++: Run time error



We cannot assign any string constant in null pointer by strcpy function.

6.
What will be output of following program?

#include<stdio.h>
int main(){
int huge *a =(int huge *)0x59990005;
int huge *b =(int huge *)0x59980015;
if(a == b)
printf("power of pointer");
else
printf("power of c");
return 0;
}
(A) power of pointer
(B) power of c
(C) power of cpower of c
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: power of pointer

Turbo C ++4.5: power of c

Linux GCC: Compilation error

Visual C++: Compilation error



Here we are performing relational operation between two huge addresses. So first of all both a and b will normalize as:

a= (0x5999)* (0x10) + (0x0005) =0x9990+0x0005=0x9995

b= (0x5998)* (0x10) + (0x0015) =0x9980+0x0015=0x9995

Here both huge addresses are representing same physical address. So a==b is true.

7.
What will be output of following program?

#include<stdio.h>
#include<string.h>
int main(){
register a = 25;
int far *p;
p=&a;
printf("%d ",*p);
return 0;
}
(A) 25
(B) 4
(C) Address
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: Compilation error

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error



Register data type stores in CPU. So it has not any memory address. Hence we cannot write &a.

8.
What will be output of following program?

#include<stdio.h>
#include<string.h>
int main(){
char far *p,*q;
printf("%d %d",sizeof(p),sizeof(q));
return 0;
}
(A) 2 2
(B) 4 4
(C) 4 2
(D) 2 4
(E) None of above


Explanation:
Turbo C++ 3.0: 4 4

Turbo C ++4.5: 4 4

Linux GCC: Compilation error

Visual C++: Compilation error



p is far pointer which size is 4 byte.

By default q is near pointer which size is 2 byte.

9.
What will be output of following program?

#include<stdio.h>
int main(){
int a = 10;
void *p = &a;
int *ptr = p;
printf("%u",*ptr);
return 0;
}
(A) 10
(B) Address
(C) 2
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: 10

Turbo C ++4.5: 10

Linux GCC: 10

Visual C++: 10



Void pointer can hold address of any data type without type casting. Any pointer can hold void pointer without type casting.

10.
What will be output of following program?

#include<stdio.h>
#include<string.h>
int main(){
int register a;
scanf("%d",&a);
printf("%d",a);
return 0;
}
//if a=25

(A) 25
(B) Address
(C) 0
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: Compilation error

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error



Register data type stores in CPU. So it has not any memory address. Hence we cannot write &a.

11.
What will be output of following program?

#include<stdio.h>
int main(){
char arr[10];
arr = "world";
printf("%s",arr);
return 0;
}
(A) world
(B) w
(C) Null
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: Compilation error

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error



Compilation error Lvalue required

Array name is constant pointer and we cannot assign any value in constant data type after declaration.

12.
What will be output of following program?

#include<stdio.h>
#include<string.h>
int main(){
int a,b,c,d;
char *p = ( char *)0;
int *q = ( int *q)0;
float *r = ( float *)0;
double *s = 0;
a = (int)(p+1);
b = (int)(q+1);
c = (int)(r+1);
d = (int)(s+1);
printf("%d %d %d %d",a,b,c,d);
return 0;
}
(A) 2 2 2 2
(B) 1 2 4 8
(C) 1 2 2 4
(D) Compilation error
(E) None of above


Explanation:



Turbo C++ 3.0: 1 2 4 8

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error



Address + 1 = next address

Since initial address of all data type is zero. So its

next address will be size of data type.

13.
What will be output of following program?

#include<stdio.h>
#include<string.h>
int main(){
int a = 5,b = 10,c;
int *p = &a,*q = &b;
c = p - q;
printf("%d" , c);
return 0;
}
(A) 1
(B) 5
(C) -5
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: 1

Turbo C ++4.5: 1

Linux GCC: 1

Visual C++: 2



Difference of two same type of pointer is always one.

14.
What will be output of following program?

#include<stdio.h>
unsigned long int (* avg())[3]{
static unsigned long int arr[3] = {1,2,3};
return &arr;
}
int main(){
unsigned long int (*ptr)[3];
ptr = avg();
printf("%d" , *(*ptr+2));
return 0;
}
(A) 1
(B) 2
(C) 3
(D) Compilation error
(E) None of above


Explanation:
Turbo C++ 3.0: 3

Turbo C ++4.5: 3

Linux GCC: 3

Visual C++: 3

15.
What will be output of following program?

#include<stdio.h>
int main(){
int * p , b;
b = sizeof(p);
printf("%d" , b);
return 0;
}
(A) 2
(B) 4
(C) 8
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: 2 or 4

Turbo C ++4.5: 2 or 4

Linux GCC: 4

Visual C++: 4



since in this question it has not written p is which type of pointer. So its output will depend upon which memory model has selected. Default memory model is small.

16.
What will be output of following program?

#include<stdio.h>
int main(){
int i = 5 , j;
int *p , *q;
p = &i;
q = &j;
j = 5;
printf("%d %d",*p,*q);
return 0;
}
(A) 5 5
(B) Address Address
(C) 5 Address
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: 5 5

Turbo C ++4.5: 5 5

Linux GCC: 5 5

Visual C++: 5 5

17.
What will be output of following program?

#include<stdio.h>
int main(){
int i = 5;
int *p;
p = &i;
printf(" %u %u", *&p , &*p);
return 0;
}
(A) 5 Address
(B) Address Address
(C) Address 5
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: Address Address

Turbo C ++4.5: Address Address

Linux GCC: Address Address

Visual C++: Address Address



Since * and & always cancel to each other.

i.e. *&a = a

so *&p = p which store address of integer i

&*p = &*(&i) //since p = &i

= &(*&i)

= &i

So second output is also address of i

18.
What will be output of following program?

#include<stdio.h>
int main(){
int i = 100;
printf("value of i : %d addresss of i : %u",i,&i);
i++;
printf("\nvalue of i : %d addresss of i : %u",i,&i);
return 0;
}

(A)
value of i : 100 addresss of i : Address
value of i : 101 addresss of i : Address
(B)
value of i : 100 addresss of i : Address
value of i : 100 addresss of i : Address
(C)
value of i : 101 addresss of i : Address
value of i : 101 addresss of i : Address
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0:

value of i : 100 addresss of i : Address

value of i : 101 addresss of i : Address   

Turbo C ++4.5:

value of i : 100 addresss of i : Address

value of i : 101 addresss of i : Address   

Linux GCC:

value of i : 100 addresss of i : Address

value of i : 101 addresss of i : Address   

Visual C++:

value of i : 100 addresss of i : Address
value of i : 101 addresss of i : Address   


Within the scope of any variable, value of variable may change but its address will never change in any modification of variable.

19.
What will be output of following program?

#include<stdio.h>  
int main(){
char far *p =(char far *)0x55550005;
char far *q =(char far *)0x53332225;
*p = 25;
(*p)++;
printf("%d",*q);
return 0;
}
(A) 25
(B) Address
(C) Garbage
(D) Compilation error
(E)None of above


Explanation:

Turbo C++ 3.0: 26

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error



Far address of p and q are representing same physical address. Physical address of

0x55550005 = 0x5555 * ox10 + ox0005 = 0x55555

Physical address of

0x53332225 = 0x5333 * 0x10 + ox2225 = 0x55555

*p = 25, means content at memory location 0x55555 is assigning value 25

(*p)++ means to increase the content by one at memory the location 0x5555 so now content of memory location at 0x55555 is 26

*q also means content at memory location 0x55555 which is 26

20.
What will be output of following program?

#include<stdio.h>  
int main(){
int i = 3;
int *j;
int **k;
j = &i;
k = &j;
printf("%u %u %u",i,j,k);
return 0;
}
(A) 3 Address 3
(B) 3 Address Address
(C) 3 3 3
(D) Compilation error
(E) None of above


Explanation:

Turbo C++ 3.0: 3 Address Address

Turbo C ++4.5: 3 Address Address

Linux GCC: 3 Address Address

Visual C++: 3 Address Address







Here 6024, 8085, 9091 is any arbitrary address, it may be different.



Pointer Tutorial

16 comments:

  1. According to Q13, the explanation is wrong. First of all, the answer in this question depends on how this 2 variables are allocated on stack, I mean if they're two consecutive values or they're apart from each other in memory. For example on my Visual C++ in such a declaration:
    int a = 5, b = 10;
    Difference between addresses of a and b is 12 bytes, so if they're 4 bytes ints the difference is &a - &b = 3.

    Difference between two pointers of same type isn't always 1 !!!!!! It depends on they relative location in memory !!!!!!

    ReplyDelete
    Replies
    1. Difference of two pointers of same type is 1 and not the size of their data types. Try this code on codepad.org which uses linux gcc compiler.

      Also, the difference between two pointers of same type is not always 1 as mentioned. Try this code

      #include
      #include
      int main(){
      int a = 5,b = 10,c = 100, d = -9;
      int *p = &a,*q = &d;
      c = p - q;
      printf("%d" , c);
      return 0;
      }

      Delete
  2. according to Q14,there is no explanation please provide explanation,it is very good for beginners...otherwise everthing fine...
    ITS VERY GOOD JOB..PLEASE KEEP IT UP

    ReplyDelete
  3. will this program face dangling pointer problem???

    #include
    unsigned long int (* avg())[3]{
    static unsigned long int arr[3] = {1,2,3};
    return &arr;
    }
    int main(){
    unsigned long int (*ptr)[3];
    ptr = avg();
    printf("%d" , *(*ptr+2));
    return 0;
    }

    ReplyDelete
  4. i want a C program that takes the input – the mark sheets of N number of students.
    for examle;
    ●Enter the marks for student 1 student name: aaa
    roll number: 24
    marks obtained in 5 subjects: 56 89 87 76 98
    ●Enter the roll number of the student to be searched: 22
    ●The details of student with roll number 22:
    name: bbb
    total marks obtained: 293

    ReplyDelete
  5. i need a C program that accepts the roll numbers and names of N students from the user … and prints those in the alphabetical order of the name, along with the roll numbers.
    Enter the number of students: 3
    Enter the name of student #1: vinay
    Enter vinay’s roll number: 12
    Enter the name of student #2: anand
    Enter anand’s roll number: 14
    Enter the name of student #3: kumar
    Enter kumar’s roll number: 11
    The list of students, in sorted order of names is given below:
    anand (14)
    kumar (11)
    vinay (12)

    ReplyDelete
  6. what is O/P of this n please also tell whats the concept behind that o/p?
    #include

    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);

    }

    ReplyDelete
    Replies
    1. should give an error as q doesn't point to a location.

      Delete
  7. when i put a=908 why i get -116 as output for program 1

    ReplyDelete
  8. ques no 2 is producing error for pointer r

    ReplyDelete
  9. there is some problem in 9 it is giving error(cant convert void to int) ..??

    ReplyDelete
  10. 7 Question output 25 in C++ 3.0

    ReplyDelete
  11. Plz help:
    Initialize an integer array, get values of array from user and store them in the array using pointer.

    ReplyDelete
  12. In Q-3, why output of **k is address in turbo c++ 4.5?give the explanataion.

    ReplyDelete
  13. In q-6, why output is power of c in turbo c++4.5? Plz explain that.

    ReplyDelete