Pointers Questions and Answers

Q: What will be the following C code print ?int a[] = {1,2,3,4};int *p = a ;printf("%d", *(++p) + 2);

The program prints 4.int *p = a; makes p point at a[0], which holds 1.++p is a pre-increment, so p moves to a[1] before it is used.*(++p) therefore reads a[1], which is 2.2 + 2 gives 4, printed with %d.3 is what you get from *p + 2 without the increment (1 + 2). 5 would need p to move two places, to a[2], which holds 3. There is no compilation error: incrementing a pointer variable is legal, and the parentheses make the increment apply to p before the dereference.

Q: In C, which of the following is TRUE about arrays and pointers ?

The true statement is that array[i] is equivalent to *(array + i).C defines subscripting in terms of pointer arithmetic: in an expression the array name becomes a pointer to its first element, adding i moves the pointer i elements along, and the dereference reads that element. The two forms are interchangeable by the language definition (which is also why the odd form i[array] compiles).sizeof(array) gives the size of the whole array, for example 40 for ten 4-byte ints, not the size of a pointer; they coincide only for an array parameter, which is really a pointer. An array name is not a modifiable lvalue, so array++ is an error. Since C is true, 'None of these' is wrong.

Q: What is the output of the following pseudocode?int arr[] = {1, 2, 3, 4, 5} int *ptr = arr + 2 print(*ptr + *(ptr + 1))

The output is 7.The array holds 1, 2, 3, 4, 5 at indices 0 to 4. arr + 2 points to index 2, so *ptr is 3. ptr + 1 points to index 3, so *(ptr + 1) is 4. The sum is 3 + 4 = 7.Pointer arithmetic moves by whole elements, not bytes, so arr + 2 is the third element. 5 is the sum of indices 1 and 2 (2 + 3), a slip of taking arr + 2 as the second element. 9 is 4 + 5, a slip of one position too far, and 3 is the first value alone.

Q: If int *p points to arr[2], then (p+1) points to

p + 1 points to arr[3].In pointer arithmetic, adding 1 to a pointer moves it forward by one element of the type it points to, not by one byte. If p holds the address of arr[2], then p + 1 holds the address of the next int, arr[3]; the compiler scales the 1 by sizeof(int) automatically.arr[1] would be p - 1. arr[2]+1 describes adding 1 to the value stored in arr[2], not moving the pointer, and the result can be determined exactly from the rules of pointer arithmetic.

Q: What is the output of the following program ?#include<stdio.h> void f(int *p, int *q) { p=q; *p=2; } int i=0, j=1; int main(void){ f(&i, &j); printf("%d %d\n", i, j); }

The program prints 0 2.C passes arguments by value, so p and q are local copies of the addresses &i and &j. Walk through f:p=q; changes only the local copy p, which now holds &j. The variable i is untouched.*p=2; writes 2 through p, that is into j.Back in main, i is still 0 and j is now 2, so printf prints 0 2. '2 1' is what you would get if *p=2 still wrote to i; '2 2' assumes p=q changed main's pointer; '0 1' ignores the write through p altogether.

Q: What is the output of the following program ?#include <stdio.h> int f (int *a, int n) { if(n<=0) return 0; else if (*a%2 = = 0) return *a + f(a+1, n-1); else return *a - f(a+1, n-1); } int main(void) { int a[] = {12, 7, 13, 4, 11, 6}; printf("%d", f(a,6)); return 0; }

The program prints 15.f adds an even element and subtracts the result of the rest after an odd element. Work from the end of the array {12, 7, 13, 4, 11, 6}:\[ \begin{aligned} f(\{6\}) &= 6 + 0 = 6 \\ f(\{11,6\}) &= 11 - 6 = 5 \\ f(\{4,\ldots\}) &= 4 + 5 = 9 \\ f(\{13,\ldots\}) &= 13 - 9 = 4 \\ f(\{7,\ldots\}) &= 7 - 4 = 3 \\ f(\{12,\ldots\}) &= 12 + 3 = 15 \end{aligned} \]The trap is reading it as an alternating sum of the elements; the subtraction applies to the whole remaining result, not to the next element alone. 5 is the value at the 11 step, and -9 and 19 come from sign slips.

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

Correct answer: 5 5 5Each dereference removes one level. k is 5; p holds the address of k, so *p is 5; and m holds the address of p, so *m is p and **m is 5. All three names reach the same object by paths of different length, which is the whole point of the double pointer.

Q: What would be the equivalent pointer expression for referring the array element a[i][j][k][l]

Correct answer: *(*(*(*(a+i)+j)+k)+l)Each subscript is defined as a dereference of a sum: x[y] means *(x+y). Applying that rule four times, from the outside in, gives one dereference for every subscript.\[ a[i][j][k][l] \equiv *(*(*(*(a+i)+j)+k)+l) \]The options without the asterisks compute addresses but never read through them. Note that the pointer arithmetic is scaled by the size of the sub-array at each level, which is why the array's dimensions must be known to the compiler.

Q: What will be the output of the program If the integer is 4bytes long? #include <stdio.h> int main() { int ***r, **q, *p, i=8; p = &i; q = &p; r = &q; printf("%d, %d, %d\n", *p, **q, ***r); return 0; }

Correct answer: 8, 8, 8Each level of indirection is undone by one dereference. p points to i, q points to p, and r points to q.So *p reaches i directly; **q goes through p to i; and ***r goes through q and p to i. All three arrive at the same object and print 8. The distractors offer addresses, which would appear only if fewer dereferences were applied than there are levels of indirection.

11.

What will be the following C code print ?

int a[] = {1,2,3,4};
int *p = a ;
printf("%d", *(++p) + 2);

A.3
B.4
C.5
D.Compilation error : invalid use of in crement operators on pointer dereference

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: B

The program prints 4.

int *p = a; makes p point at a[0], which holds 1.
++p is a pre-increment, so p moves to a[1] before it is used.
*(++p) therefore reads a[1], which is 2.
2 + 2 gives 4, printed with %d.

3 is what you get from *p + 2 without the increment (1 + 2). 5 would need p to move two places, to a[2], which holds 3. There is no compilation error: incrementing a pointer variable is legal, and the parentheses make the increment apply to p before the dereference.

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12.

In C, which of the following is TRUE about arrays and pointers ?

A.sizeof(array) equals sizeof(pointer)
B.Array name can be incremented
C.array[i] is equivalent to *(array + i)
D.None of these

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: C

The true statement is that array[i] is equivalent to *(array + i).

C defines subscripting in terms of pointer arithmetic: in an expression the array name becomes a pointer to its first element, adding i moves the pointer i elements along, and the dereference reads that element. The two forms are interchangeable by the language definition (which is also why the odd form i[array] compiles).

sizeof(array) gives the size of the whole array, for example 40 for ten 4-byte ints, not the size of a pointer; they coincide only for an array parameter, which is really a pointer. An array name is not a modifiable lvalue, so array++ is an error. Since C is true, 'None of these' is wrong.

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13.

What is the output of the following pseudocode?

int arr[] = {1, 2, 3, 4, 5}
int *ptr = arr + 2
print(*ptr + *(ptr + 1))

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: C

The output is 7.

The array holds 1, 2, 3, 4, 5 at indices 0 to 4. arr + 2 points to index 2, so *ptr is 3. ptr + 1 points to index 3, so *(ptr + 1) is 4. The sum is 3 + 4 = 7.

Pointer arithmetic moves by whole elements, not bytes, so arr + 2 is the third element. 5 is the sum of indices 1 and 2 (2 + 3), a slip of taking arr + 2 as the second element. 9 is 4 + 5, a slip of one position too far, and 3 is the first value alone.

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14.

If int *p points to arr[2], then (p+1) points to

A.arr[1]
B.arr[3]
C.arr[2]+1
D.Cannot determine

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: B

p + 1 points to arr[3].

In pointer arithmetic, adding 1 to a pointer moves it forward by one element of the type it points to, not by one byte. If p holds the address of arr[2], then p + 1 holds the address of the next int, arr[3]; the compiler scales the 1 by sizeof(int) automatically.

arr[1] would be p - 1. arr[2]+1 describes adding 1 to the value stored in arr[2], not moving the pointer, and the result can be determined exactly from the rules of pointer arithmetic.

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15.

What is the output of the following program ?

#include<stdio.h>
void f(int *p, int *q) {
    p=q;
    *p=2;
}
int i=0, j=1;
int main(void){
    f(&i, &j);
    printf("%d %d\n", i, j);
    }

A.2 2
B.2 1
C.0 1
D.0 2

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: D

The program prints 0 2.

C passes arguments by value, so p and q are local copies of the addresses &i and &j. Walk through f:

p=q; changes only the local copy p, which now holds &j. The variable i is untouched.
*p=2; writes 2 through p, that is into j.

Back in main, i is still 0 and j is now 2, so printf prints 0 2. '2 1' is what you would get if *p=2 still wrote to i; '2 2' assumes p=q changed main's pointer; '0 1' ignores the write through p altogether.

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16.

What is the output of the following program ?

#include <stdio.h>
int f (int *a, int n)
{
if(n<=0) return 0;
else if (*a%2 = = 0) return *a + f(a+1, n-1);
else return *a - f(a+1, n-1);
}
int main(void)
{
int a[] = {12, 7, 13, 4, 11, 6};
printf("%d", f(a,6));
return 0;
}

A.-9
B.5
C.15
D.19

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: C

The program prints 15.

f adds an even element and subtracts the result of the rest after an odd element. Work from the end of the array {12, 7, 13, 4, 11, 6}:

\[ \begin{aligned} f(\{6\}) &= 6 + 0 = 6 \\ f(\{11,6\}) &= 11 - 6 = 5 \\ f(\{4,\ldots\}) &= 4 + 5 = 9 \\ f(\{13,\ldots\}) &= 13 - 9 = 4 \\ f(\{7,\ldots\}) &= 7 - 4 = 3 \\ f(\{12,\ldots\}) &= 12 + 3 = 15 \end{aligned} \]

The trap is reading it as an alternating sum of the elements; the subtraction applies to the whole remaining result, not to the next element alone. 5 is the value at the 11 step, and -9 and 19 come from sign slips.

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17.

What does the following declaration mean ? int (*ptr) [10];

A.ptr is an array of pointers of 10 integers.
B.ptr is a pointer to an array of 10 integers.
C.ptr is an array of 10 integers.
D.none of the above.

Exam Year(s)

UGC NET - 2013,

Answer: B

The declaration int (*ptr)[10]; means that ptr is a pointer to an array of 10 integers. The parentheses bind the star to ptr first, so ptr is a pointer, and the [10] then describes the thing it points to.

An array of 10 pointers to integers would be written int *ptr[10], without the parentheses. A plain array of 10 integers would be int ptr[10]. Since the declaration is a valid one with a clear meaning, 'none of the above' is wrong.

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18.

What is the output of this C code?

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

A.5 5 5
B.5 5 junk value
C.5 junk value
D.Run time error

Answer: A

Correct answer: 5 5 5

Each dereference removes one level. k is 5; p holds the address of k, so *p is 5; and m holds the address of p, so *m is p and **m is 5. All three names reach the same object by paths of different length, which is the whole point of the double pointer.

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19.

What would be the equivalent pointer expression for referring the array element a[i][j][k][l]

A.((((a+i)+j)+k)+l)
B.*(*(*(*(a+i)+j)+k)+l)
C.(((a+i)+j)+k+l)
D.((a+i)+j+k+l)

Answer: B

Correct answer: *(*(*(*(a+i)+j)+k)+l)

Each subscript is defined as a dereference of a sum: x[y] means *(x+y). Applying that rule four times, from the outside in, gives one dereference for every subscript.

\[ a[i][j][k][l] \equiv *(*(*(*(a+i)+j)+k)+l) \]

The options without the asterisks compute addresses but never read through them. Note that the pointer arithmetic is scaled by the size of the sub-array at each level, which is why the array's dimensions must be known to the compiler.

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20.

What will be the output of the program If the integer is 4bytes long?

#include <stdio.h>

int main()
{
    int ***r, **q, *p, i=8;
    p = &i;
    q = &p;
    r = &q;
    printf("%d, %d, %d\n", *p, **q, ***r);
    return 0;
}

A.8, 8, 8
B.4000, 4002, 4004
C.4000, 4004, 4008
D.4000, 4008, 4016

Answer: A

Correct answer: 8, 8, 8

Each level of indirection is undone by one dereference. p points to i, q points to p, and r points to q.

So *p reaches i directly; **q goes through p to i; and ***r goes through q and p to i. All three arrive at the same object and print 8. The distractors offer addresses, which would appear only if fewer dereferences were applied than there are levels of indirection.

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