Official PYQ · 2015Set 3 · Q48Programming in CNAT1 mark

GATE CS 2015 Set 3 Q48 — Programming in C

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 _______.

View worked solution

Answer: 10

To find the number of times the printf statement is executed, we trace the program step-by-step:

Step 1: Calculate the initial values of $j$ and $k$

2 * 3 / 4 evaluates to 6 / 4 = 1 (integer division).
2.0 / 5 evaluates to 0.4 (floating-point division).
8 / 5 evaluates to 1 (integer division).
Thus, $j = 1 + 0.4 + 1 = 2.4$ . Since $j$ is of type int, it is truncated to $2$ .
Next, k -= --j decrements $j$ to $1$ first, then $k = 0 - 1 = -1$ .

Step 2: Trace the loop for $i = 0$ to $4$

The switch expression is $i + k = i - 1$ . Since there are no break statements, execution falls through to subsequent cases once a match is found:

$i$	$i + k$	Matching Case & Fall-through Path	`printf` Executions
0	-1	`default`	1
1	0	`default`	1
2	1	`case 1` → `case 2` → `case 3` → `default`	3
3	2	`case 2` → `case 3` → `default`	3
4	3	`case 3` → `default`	2

Total Executions: $1 + 1 + 3 + 3 + 2 = 10$ .

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Published 10 Oct 2026

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