GATE CS 2017 Set 2 — Question 14

MCQ+1 / -0.33EasyPointers & Pointer ArithmeticC ProgrammingProgramming & Data Structures

Programming & Data Structures → C Programming → Pointers & Pointer Arithmetic

Last updated 5 September 2026

Question

Consider the following function implemented in C: void printxy(int x, int y) { int *ptr; x = 0; ptr = &x; y = *ptr; *ptr = 1; printf("%d, %d", x, y); } The output of invoking printxy(1, 1) is

Correct answer (C) 1, 0 Solution Let's trace the execution of printxy(1, 1) : 1. Initial values: x = 1, y = 1 . 2. x = 0; -> x becomes 0. 3. ptr = &x; -> ptr now stores the address of x . 4. y = *ptr; -> y is assigned the value at the address stored in ptr (which is x). So, y = 0 . 5. *ptr = 1; -> The value at the address stored in ptr (which is x) is set to 1. So, x = 1 . 6. printf("%d, %d", x, y); -> Prints the current values of x and y, which are 1 and 0 respectively. Thus, the output is 1, 0 . Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer (C) 1, 0 Solution Let's trace the execution of printxy(1, 1) : 1. Initial values: x = 1, y = 1 . 2. x = 0; -> x becomes 0. 3. ptr = &x; -> ptr now stores the address of x . 4. y = *ptr; -> y is assigned the value at the address stored in ptr (which is x). So, y = 0 . 5. *ptr = 1; -> The value at the address stored in ptr (which is x) is set to 1. So, x = 1 . 6. printf("%d, %d", x, y); -> Prints the current values of x and y, which are 1 and 0 respectively. Thus, the output is 1, 0 .

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