What will be output of following code?

Question 01:

#include <stdio.h>
int main()
{
    int i = 0;
    for(i = 0; i < 20; i++) {
        switch(i) {
            case 0: i += 5;
            case 1: i += 2;
            case 5: i += 5;
            default: i += 4;
                     break;
        }
    }
    printf("%d", i);
}

Let’s analyze how the code works step-by-step.

The initial setup:

• i starts at 0.
• The loop runs for(i = 0; i < 20; i++).

Within each iteration of the for loop:

• The switch statement executes, checking the current value of i.
• The cases inside the switch don't use break statements until default, so the cases "fall through," meaning once a case is matched, subsequent cases are also executed until a break or the end of the switch.

Let’s go through each iteration in detail:

Step-by-Step Execution

Iteration 1: i = 0

• Switch starts with case 0 because i = 0.
• i += 5 (so i becomes 5).
• Falls through to case 1, so i += 2 (making i = 7).
• Falls through to case 5, so i += 5 (making i = 12).
• Falls through to default, so i += 4 (making i = 16).
• Break statement in default exits the switch.

Iteration 2: i = 17 (incremented by for loop)

Switch starts with default (since i does not match any case).

• i += 4 (making i = 21).
• Break statement in default exits the switch.

Loop ends: Now i = 22, which is greater than 20, so the loop stops.

The final value of i is 22, so the program prints 22.

Question 02:

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Let’s go through the code step-by-step to understand why it outputs 1.

#include <stdio.h>
int main()
{
    int i = 2, j = 2;
    while(i + 1 ? --i : j++) {
        printf("%d", i);
    }
}

Explanation of the Code

1. Variable Initialization: i and j are both initialized to 2.
2. While Loop Condition: i + 1 ? --i : j++

• The loop condition uses the ternary operator (condition ? expr_if_true : expr_if_false).
• The expression i + 1 is evaluated first:
• Since i = 2, i + 1 becomes 3, which is non-zero (truthy).
• Because of this, the true branch (--i) is executed instead of j++.
• The while loop will continue as long as the result of the ternary operator is non-zero.

1. Loop Execution:

• Let’s go through each iteration of the loop to see how i changes.

Step-by-Step Execution

1. First Iteration:

• i = 2
• i + 1 is 3, which is truthy, so --i is executed.
• i becomes 1 (--i decrements i by 1).
• printf("%d", i); prints 1.

2. Second Iteration:

• i = 1
• i + 1 is 2, which is still truthy, so --i is executed again.
• i becomes 0.
• Since printf("%d", i); is inside the loop body, it does not execute in this iteration because the loop condition evaluates to 0 (falsy) after the decrement.

3. Loop Exit:

• The loop exits because i + 1 becomes 0 when i is 0, which makes the ternary condition false.
• No further values are printed.

The only value printed by printf is 1.

Question 03:

#include <stdio.h>
int main()
{
    int i, j;
    i = j = 2, 3;
    while (--i && j++) {
        printf("%d - %d", i, j);
    }
}

Explanation of the Code

Variable Initialization:

• i = j = 2, 3;
• This line might seem confusing because of the comma.
• i = j = 2 is executed first due to operator precedence, which sets both i and j to 2.
• The , 3 part is simply ignored because it has no effect here; it's an isolated expression after the assignments.

So, i and j are both initialized to 2.

While Loop:

• while (--i && j++)
• The loop condition has two expressions: --i and j++.
• The && (logical AND) operator means the loop will continue as long as both --i and j++ evaluate to non-zero (true).

Let's go through each iteration of the loop.

Step-by-Step Execution

First Iteration:

• --i: i is decremented from 2 to 1.
• j++: j is currently 2, but after j++, it increments to 3.
• Since --i is 1 (true) and j++ was 2 (also true), the loop condition is satisfied, and the loop body executes.
• printf("%d - %d", i, j); prints the current values of i and j, which are 1 and 3, respectively.

Second Iteration:

--i: i is decremented from 1 to 0.

Since --i is now 0 (false), the loop condition --i && j++ fails, and the loop exits without printing anything further.

Output: 1 – 3