GATE CS 2015 Set 3 — Question 53

NAT+2 / -0MediumData Types & OperatorsC ProgrammingProgramming & Data StructuresControl Flow

Programming & Data Structures → C Programming → Control Flow

Last updated 5 September 2026

Question

Consider the following C program:#include<stdio.h> int main( ) { int i, j, k = 0; j = 2 * 3 / 4 + 2.0 / 5 + 8 / 5; k -= --j; for(i = 0; i < 5; i++) { switch(i + k) { case 1: case 2: printf("\n%d", i+k); case 3: printf("\n%d", i+k); default: printf("\n%d", i+k); } } return 0; } The number of times printf statement is executed is __________.

Correct answer 10 to 10 Solution Step 1: Evaluate j . j = 2 * 3 / 4 + 2.0 / 5 + 8 / 5 2 * 3 = 6, 6 / 4 = 1 (integer division) 2.0 / 5 = 0.4 (float division) 8 / 5 = 1 (integer division) j = 1 + 0.4 + 1 = 2.4 Since j is an integer, j = 2 . Step 2: Evaluate k . k -= --j Pre-decrement j becomes 1. k = k - 1 = 0 - 1 = -1 . Step 3: Loop i from 0 to 4. Inside loop, switch on i + k (which is i - 1). i = 0 : switch(-1) . No match. Goes to default . Prints once. i = 1 : switch(0) . No match. Goes to default . Prints once. i = 2 : switch(1) . Matches case 1 . Falls through to case 2 (print), case 3 (print), default (print). Total 3 prints. i = 3 : switch(2) . Matches case 2 . Prints. Falls through to case 3 (print), default (print). Total 3 prints. i = 4 : switch(3) . Matches case 3 . Prints. Falls through to default (print). Total 2 prints. Total executions = 1 + 1 + 3 + 3 + 2 = 10. Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer 10 to 10 Solution Step 1: Evaluate j . j = 2 * 3 / 4 + 2.0 / 5 + 8 / 5 2 * 3 = 6, 6 / 4 = 1 (integer division) 2.0 / 5 = 0.4 (float division) 8 / 5 = 1 (integer division) j = 1 + 0.4 + 1 = 2.4 Since j is an integer, j = 2 . Step 2: Evaluate k . k -= --j Pre-decrement j becomes 1. k = k - 1 = 0 - 1 = -1 . Step 3: Loop i from 0 to 4. Inside loop, switch on i + k (which is i - 1). i = 0 : switch(-1) . No match. Goes to default . Prints once. i = 1 : switch(0) . No match. Goes to default . Prints once. i = 2 : switch(1) . Matches case 1 . Falls through to case 2 (print), case 3 (print), default (print). Total 3 prints. i = 3 : switch(2) . Matches case 2 . Prints. Falls through to case 3 (print), default (print). Total 3 prints. i = 4 : switch(3) . Matches case 3 . Prints. Falls through to default (print). Total 2 prints. Total executions = 1 + 1 + 3 + 3 + 2 = 10.

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