GATE CS 2015 Set 3 — Question 43

NAT+2 / -0MediumPointers & Pointer ArithmeticC ProgrammingProgramming & Data StructuresArrays & Strings in C

Programming & Data Structures → C Programming → Arrays & Strings in C

Last updated 5 September 2026

Question

Consider the following C program.#include<stdio.h> int main( ) { static int a[ ] = {10, 20, 30, 40, 50}; static int *p[ ] = {a, a+3, a+4, a+1, a+2}; int **ptr = p; ptr++; printf("%d%d", ptr-p,**ptr); } The output of the program is ___________.

Correct answer 140 to 140 Solution Let's analyze the code step by step: 1. static int a[] = {10, 20, 30, 40, 50}; Array a has elements at indices 0 to 4. Let address of a be 1000 (assuming int size 4). a[0] at 1000, a[1] at 1004, etc. 2. static int *p[] = {a, a+3, a+4, a+1, a+2}; p is an array of pointers. p[0] = a (points to a[0] = 10) p[1] = a+3 (points to a[3] = 40) p[2] = a+4 (points to a[4] = 50) p[3] = a+1 (points to a[1] = 20) p[4] = a+2 (points to a[2] = 30) 3. int **ptr = p; ptr points to the first element of array p, i.e., ptr = &p[0] . 4. ptr++; ptr is incremented to point to the next element of p . Now ptr = &p[1] . 5. printf("%d%d", ptr-p, **ptr); ptr - p : Pointer arithmetic. ptr is &p[1], p is &p[0] . The difference is 1 element. So this prints 1 . **ptr : *ptr gives the value ptr points to, which is p[1] . p[1] is the address a+3 . **ptr dereferences a+3, giving a[3], which is 40 . So this prints 40 . The combined output is 140 . Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer 140 to 140 Solution Let's analyze the code step by step: 1. static int a[] = {10, 20, 30, 40, 50}; Array a has elements at indices 0 to 4. Let address of a be 1000 (assuming int size 4). a[0] at 1000, a[1] at 1004, etc. 2. static int *p[] = {a, a+3, a+4, a+1, a+2}; p is an array of pointers. p[0] = a (points to a[0] = 10) p[1] = a+3 (points to a[3] = 40) p[2] = a+4 (points to a[4] = 50) p[3] = a+1 (points to a[1] = 20) p[4] = a+2 (points to a[2] = 30) 3. int **ptr = p; ptr points to the first element of array p, i.e., ptr = &p[0] . 4. ptr++; ptr is incremented to point to the next element of p . Now ptr = &p[1] . 5. printf("%d%d", ptr-p, **ptr); ptr - p : Pointer arithmetic. ptr is &p[1], p is &p[0] . The difference is 1 element. So this prints 1 . **ptr : *ptr gives the value ptr points to, which is p[1] . p[1] is the address a+3 . **ptr dereferences a+3, giving a[3], which is 40 . So this prints 40 . The combined output is 140 .

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