GATE CS 2017 Set 1 — Question 55

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Programming & Data Structures → C Programming → Data Types & Operators

Last updated 5 September 2026

Question

The output of executing the following C program is __________.#include <stdio.h> int total(int v) { static int count = 0; while(v) { count += v&1; v >>= 1; } return count; } void main() { static int x = 0; int i = 5; for(; i > 0; i--) { x = x + total(i); } printf("%d\n", x); }

Correct answer 23 to 23 Solution Let's trace the execution of the program. The key is to note that both the count variable in the total function and the x variable in the main function are declared as static . This means they retain their values between function calls and iterations. count in total : Initialized to 0 once. Its value is cumulative across all calls to total . x in main : Initialized to 0 once. The for loop in main iterates for i = 5, 4, 3, 2, 1 . 1. i = 5: total(5) is called. v = 5 (binary 101). count is initially 0. The while loop in total counts the set bits in v and adds them to the static count . Number of set bits in 5 is 2. count becomes 0 + 2 = 2 . total returns the new value of count, which is 2. In main, x = x + total(5) becomes x = 0 + 2 = 2 . 2. i = 4: total(4) is called. v = 4 (binary 100). count is currently 2. Number of set bits in 4 is 1. count becomes 2 + 1 = 3 . total returns 3. In main, x = x + total(4) becomes x = 2 + 3 = 5 . 3. i = 3: total(3) is called. v = 3 (binary 11). count is currently 3. Number of set bits in 3 is 2. count becomes 3 + 2 = 5 . total returns 5. In main, x = x + total(3) becomes x = 5 + 5 = 10 . 4. i = 2: total(2) is called. v = 2 (binary 10). count is currently 5. Number of set bits in 2 is 1. count becomes 5 + 1 = 6 . total returns 6. In main, x = x + total(2) becomes x = 10 + 6 = 16 . 5. i = 1: total(1) is called. v = 1 (binary 1). count is currently 6. Number of set bits in 1 is 1. count becomes 6 + 1 = 7 . total returns 7. In main, x = x + total(1) becomes x = 16 + 7 = 23 . After the loop finishes, the printf statement prints the final value of x, which is 23. Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer 23 to 23 Solution Let's trace the execution of the program. The key is to note that both the count variable in the total function and the x variable in the main function are declared as static . This means they retain their values between function calls and iterations. count in total : Initialized to 0 once. Its value is cumulative across all calls to total . x in main : Initialized to 0 once. The for loop in main iterates for i = 5, 4, 3, 2, 1 . 1. i = 5: total(5) is called. v = 5 (binary 101). count is initially 0. The while loop in total counts the set bits in v and adds them to the static count . Number of set bits in 5 is 2. count becomes 0 + 2 = 2 . total returns the new value of count, which is 2. In main, x = x + total(5) becomes x = 0 + 2 = 2 . 2. i = 4: total(4) is called. v = 4 (binary 100). count is currently 2. Number of set bits in 4 is 1. count becomes 2 + 1 = 3 . total returns 3. In main, x = x + total(4) becomes x = 2 + 3 = 5 . 3. i = 3: total(3) is called. v = 3 (binary 11). count is currently 3. Number of set bits in 3 is 2. count becomes 3 + 2 = 5 . total returns 5. In main, x = x + total(3) becomes x = 5 + 5 = 10 . 4. i = 2: total(2) is called. v = 2 (binary 10). count is currently 5. Number of set bits in 2 is 1. count becomes 5 + 1 = 6 . total returns 6. In main, x = x + total(2) becomes x = 10 + 6 = 16 . 5. i = 1: total(1) is called. v = 1 (binary 1). count is currently 6. Number of set bits in 1 is 1. count becomes 6 + 1 = 7 . total returns 7. In main, x = x + total(1) becomes x = 16 + 7 = 23 . After the loop finishes, the printf statement prints the final value of x, which is 23.

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GATE CS 2017 Set 1 — Question 55

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