[Solved] Identify the error in this code snippet using continue: int i = 0; while (i — Ans: The loop will run infinitely because 'i' is not incremented when i == 2 | C

C - Loop Control Statements

Identify the error in this code snippet using continue:

int i = 0;
while (i < 5) {
    if (i == 2)
        continue;
    printf("%d ", i);
    i++;
}

ASyntax error: continue cannot be used in while loops

BThe printf statement is unreachable

CThe loop will run infinitely because 'i' is not incremented when i == 2

DThe variable 'i' is not declared properly

Step-by-Step Solution

Solution:

Step 1: Understand loop flow with continue
When i equals 2, continue skips the rest of the loop body, so i++ is not executed.
Step 2: Consequence of skipping i++
Since i is not incremented at i == 2, the loop condition remains true forever, causing an infinite loop.
Final Answer:
The loop will run infinitely because 'i' is not incremented when i == 2 -> Option C
Quick Check:
continue skips i++ causing infinite loop [OK]

Quick Trick: Always increment loop variable before continue to avoid infinite loops [OK]

Common Mistakes:

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