C++ Institute Exam CPA-21-02 Topic 1 Question 39 Discussion

Actual exam question for C++ Institute's CPA-21-02 exam

Question #: 39
Topic #: 1

[All CPA-21-02 Questions]

What happens when you attempt to compile and run the following code?

#include

using namespace std;

int main(){

int i = 1;

if (i++==1) {

cout << i;

} else {

cout << i-1;

}

return 0;

}

AIt prints: 0

BIt prints: 1

CIt prints: -1

DIt prints: 2

Show Suggested Answer

Suggested Answer: D

by Teri at May 02, 2025, 10:59 AM

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Basilia

2 months ago

Haha, this is a classic trick question! The right answer is B, because the post-increment operator `i++` returns the original value of `i`, which is 1, and then increments it. So the output is 2, not 1.

upvoted 0 times

Nohemi

14 days ago

User1: Got it, thanks for clarifying!

upvoted 0 times

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Denny

15 days ago

User3: Oh, I see! Thanks for the explanation.

upvoted 0 times

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Julio

27 days ago

User2: Actually, it prints: 2

upvoted 0 times

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Jillian

1 months ago

User1: I think it prints: 1

upvoted 0 times

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Regenia

2 months ago

I think the correct answer is B. The `if (i++==1)` condition evaluates to true, and the `cout << i;` statement prints the incremented value of `i`, which is 2.

upvoted 0 times

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