Spot the effect of a stray semicolon after a <code>for</code> loop in C. What does this program print? #include<stdio.h> int main() { int i = 0; for (; i <= 5; i++); printf("%d", i); return 0; }

Introduction / Context:
This exercise highlights a common bug: placing a semicolon immediately after a for loop header. That semicolon makes the loop body empty, and the following statement executes once, after the loop finishes.

Given Data / Assumptions:

• Loop header: for (; i <= 5; i++); has an empty body.
• After the loop, a single printf prints the current value of i.

Concept / Approach:
The loop increments i from 0 up to 6, stopping when the condition i <= 5 becomes false. The moment the loop terminates, i holds 6. Then, and only then, the printf executes once and prints the final value of i with no newline.

Step-by-Step Solution:

Verification / Alternative check:
Add braces to make the intended body explicit or remove the stray semicolon to iterate printing multiple numbers.

Why Other Options Are Wrong:

Common Pitfalls:
Misplacing semicolons after loop headers; forgetting braces for multi-statement bodies.

Final Answer:
6.