Pointers Questions and Answers

Q: What is (void*)0?

Correct answer: Representation of NULL pointer(void*)0 is an integer constant expression of value zero cast to void *, which is precisely the definition of a null pointer constant. The NULL macro is typically defined as exactly this.It is not merely "a void pointer" - the value matters as much as the type. Note also that the null pointer's internal representation need not be all-zero bits on every machine; what the standard guarantees is that this constant converts to a null pointer of any type.

Q: What will be the output of the program ? #include <stdio.h> #include <string.h> int main() { int i, n; char *x="Alice"; n = strlen(x); *x = x[n]; for(i=0; i<=n; i++) { printf("%s ", x); x++; } printf("\n", x); return 0; }

Correct answer: lice ice ce ex[n] is the terminating null, and *x = x[n] writes it over the first character, so the string becomes an empty string followed by lice.The first iteration therefore prints nothing before its space. Each subsequent iteration advances x by one and prints the remaining tail - lice, ice, ce, e. Note that x points to a string literal, so writing through it is undefined behaviour and would fault on many modern systems.

Q: Which of the following does not initialize ptr to null (assuming variable declaration of a as int a=0)?

Correct answer: int *ptr = &a;That initialiser gives ptr the address of a, which is a real object and therefore not a null pointer.The other two do yield null. &a - &a is zero, and a - a is zero because a is 0 - and a constant expression evaluating to 0 is a null pointer constant, so it converts. Both are legal but poor style: NULL states the intention, and a subtraction that happens to give zero conceals it.

Q: What is the output of this C code? void main() { int k = 5; int *p = &k; int **m = &p; **m = 6; printf("%d\n", k); }

Correct answer: 6**m reaches all the way through to k: m points at p, and p points at k. So the assignment writes 6 into k itself, and printing k confirms it. This is how a double pointer lets code several levels away modify the original object.

Q: What is the output of this C code? int main() { int a = 1, b = 2, c = 3; int *ptr1 = &a, *ptr2 = &b, *ptr3 = &c; int **sptr = &ptr1; //-Ref *sptr = ptr2; }

Correct answer: ptr1 points to bsptr holds the address of ptr1, so *sptr is ptr1, and assigning ptr2 to it copies the address of b into ptr1. Afterwards ptr1 and ptr2 both point at b. Note that sptr itself is unchanged and still points at ptr1.

Q: What will be the output of the program ? #include <stdio.h> int main() { char *str; str = "%s"; printf(str, "K\n"); return 0; }

Correct answer: Kprintf takes its format as an ordinary const char * argument, so nothing requires it to be a literal written at the call site. Here str holds "%s", and printf(str, "K\n") uses it exactly as if the literal had been written there.The output is K followed by a newline. The technique is legal but dangerous in practice: a format string taken from input rather than fixed in the source is the classic format string vulnerability, which is why compilers warn about non-literal formats.

Q: What will be the output of the program assuming that the array begins at location 1002? #include <stdio.h> int main() { int a[2][3][4] = { {1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 2}, {2, 1, 4, 7, 6, 7, 8, 9, 0, 0, 0, 0} }; printf("%u, %u, %u, %d\n", a, *a, **a, ***a); return 0; }

Correct answer: 1002, 1002, 1002, 1The first three arguments all denote the same address, differing only in type. a is a pointer to the first two-dimensional sub-array, *a a pointer to its first row, and **a a pointer to its first element - but the first element sits at the start of the first row, which sits at the start of the array, so all three are 1002.Only ***a dereferences all the way through to a value, giving the first element, 1. This is the standard demonstration that an array's address and its first element's address coincide numerically while differing in type.

Q: The operator used to get value at address stored in a pointer variable is

Correct answer: *The unary * is the indirection, or dereference, operator: it yields the object a pointer points to. Its counterpart & is the address-of operator, going the other way.The asterisk plays two distinct roles that are easy to conflate. In a declaration it forms a pointer type; in an expression it dereferences. The && and || operators are logical AND and OR and have nothing to do with pointers.

Q: What is the output of this C code? int main() { int i = 10; void *p = &i; printf("%d\n", (int)*p); return 0; }

Correct answer: Compile time errorA void * cannot be dereferenced. void is an incomplete type with no size and no representation, so the compiler cannot know how many bytes to read or how to interpret them, and *p is rejected before the cast is even considered. The correct form casts the pointer first: *(int *)p.

31.

What is (void*)0?

A.Representation of NULL pointer
B.Representation of void pointer
C.Error
D.None of above

Answer: A

Correct answer: Representation of NULL pointer

(void*)0 is an integer constant expression of value zero cast to void *, which is precisely the definition of a null pointer constant. The NULL macro is typically defined as exactly this.

It is not merely "a void pointer" - the value matters as much as the type. Note also that the null pointer's internal representation need not be all-zero bits on every machine; what the standard guarantees is that this constant converts to a null pointer of any type.

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32.

What will be the output of the program ?

#include <stdio.h>
#include <string.h>

int main()
{
    int i, n;
    char *x="Alice";
    n = strlen(x);
    *x = x[n];
    for(i=0; i<=n; i++)
    {
        printf("%s ", x);
        x++;
    }
    printf("\n", x);
    return 0;
}

A.Alice
B.ecilA
C.Alice lice ice ce e
D.lice ice ce e

Answer: D

Correct answer: lice ice ce e

x[n] is the terminating null, and *x = x[n] writes it over the first character, so the string becomes an empty string followed by lice.

The first iteration therefore prints nothing before its space. Each subsequent iteration advances x by one and prints the remaining tail - lice, ice, ce, e. Note that x points to a string literal, so writing through it is undefined behaviour and would fault on many modern systems.

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33.

Which of the following does not initialize ptr to null (assuming variable declaration of a as int a=0)?

A.int *ptr = &a;
B.int *ptr = &a - &a;
C.int *ptr = a - a;
D.All of the mentioned

Answer: A

Correct answer: int *ptr = &a;

That initialiser gives ptr the address of a, which is a real object and therefore not a null pointer.

The other two do yield null. &a - &a is zero, and a - a is zero because a is 0 - and a constant expression evaluating to 0 is a null pointer constant, so it converts. Both are legal but poor style: NULL states the intention, and a subtraction that happens to give zero conceals it.

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34.

What is the output of this C code?

void main()
{
int k = 5;
int *p = &k;
int **m = &p;
**m = 6;
printf("%d\n", k);
}

A.5
B.Compile time error
C.6
D.Junk

Answer: C

Correct answer: 6

**m reaches all the way through to k: m points at p, and p points at k. So the assignment writes 6 into k itself, and printing k confirms it. This is how a double pointer lets code several levels away modify the original object.

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35.

What is the output of this C code?

int main()
{
int a = 1, b = 2, c = 3;
int *ptr1 = &a, *ptr2 = &b, *ptr3 = &c;
int **sptr = &ptr1; //-Ref
*sptr = ptr2;
}

A.ptr1 points to a
B.ptr1 points to b
C.sptr points to ptr2
D.None of the mentioned

Answer: B

Correct answer: ptr1 points to b

sptr holds the address of ptr1, so *sptr is ptr1, and assigning ptr2 to it copies the address of b into ptr1. Afterwards ptr1 and ptr2 both point at b. Note that sptr itself is unchanged and still points at ptr1.

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36.

What will be the output of the program ?

#include <stdio.h>

int main()
{
    char *str;
    str = "%s";
    printf(str, "K\n");
    return 0;
}

A.Error
B.No output
C.K
D.%s

Answer: C

Correct answer: K

printf takes its format as an ordinary const char * argument, so nothing requires it to be a literal written at the call site. Here str holds "%s", and printf(str, "K\n") uses it exactly as if the literal had been written there.

The output is K followed by a newline. The technique is legal but dangerous in practice: a format string taken from input rather than fixed in the source is the classic format string vulnerability, which is why compilers warn about non-literal formats.

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37.

What will be the output of the program assuming that the array begins at location 1002?

#include <stdio.h>

int main()
{
    int a[2][3][4] = { {1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 2}, 
                       {2, 1, 4, 7, 6, 7, 8, 9, 0, 0, 0, 0} };
    printf("%u, %u, %u, %d\n", a, *a, **a, ***a);
    return 0;
}

A.1002, 2004, 4008, 2
B.2004, 4008, 8016, 1
C.1002, 1002, 1002, 1
D.Error

Answer: C

Correct answer: 1002, 1002, 1002, 1

The first three arguments all denote the same address, differing only in type. a is a pointer to the first two-dimensional sub-array, *a a pointer to its first row, and **a a pointer to its first element - but the first element sits at the start of the first row, which sits at the start of the array, so all three are 1002.

Only ***a dereferences all the way through to a value, giving the first element, 1. This is the standard demonstration that an array's address and its first element's address coincide numerically while differing in type.

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38.

The operator used to get value at address stored in a pointer variable is

A.*
B.&
C.&&
D.||

Answer: A

Correct answer: *

The unary * is the indirection, or dereference, operator: it yields the object a pointer points to. Its counterpart & is the address-of operator, going the other way.

The asterisk plays two distinct roles that are easy to conflate. In a declaration it forms a pointer type; in an expression it dereferences. The && and || operators are logical AND and OR and have nothing to do with pointers.

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39.

What is the output of this C code?

int main()
{
int i = 10;
void *p = &i;
printf("%d\n", (int)*p);
return 0;
}

A.Compile time error
B.Segmentation fault/runtime crash
C.10
D.Undefined behaviour

Answer: A

Correct answer: Compile time error

A void * cannot be dereferenced. void is an incomplete type with no size and no representation, so the compiler cannot know how many bytes to read or how to interpret them, and *p is rejected before the cast is even considered. The correct form casts the pointer first: *(int *)p.

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40.

Comment on the output of this C code?

int main()
{
int a = 10;
int **c -= &&a;
}

A.You cannot apply any arithmetic operand to a pointer.
B.We don’t have address of an address operator
C.Both (a) and (b)
D.None of the mentioned.

Answer: B

Correct answer: We don't have address of an address operator

There is no && address operator in C. The token exists only as logical AND, so &&a is parsed as that operator with a missing left operand and the line will not compile. To obtain a pointer to a pointer you must have a pointer object to take the address of, as the earlier questions do with &p.

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