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

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
D) It prints: 2
A) It prints: 0
B) It prints: 1
C) It prints: -1

C++ Institute CPA-21-02 Exam - 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;

}

Show Suggested Answer Hide Answer
Suggested Answer: D

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

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Mindy
8 months ago
Nope, it’s not 0 or -1 either.
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Galen
9 months ago
I thought it would print 1, but I guess not!
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Loren
9 months ago
Totally agree, it’s definitely 2!
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Tonette
9 months ago
It prints: 2
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Chantay
9 months ago
Wait, how does it end up being 2?
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Antonio
9 months ago
Wait, could it be that it prints 1? I feel like the increment happens after the comparison, so I'm confused about the output.
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Noemi
10 months ago
I practiced a similar question, and I think it should print 2 because `i` is incremented after the condition is evaluated.
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Magda
10 months ago
I'm not entirely sure, but I remember something about post-increment. Maybe it prints 1 because `i` is still 1 when checked?
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Ressie
10 months ago
I think the code checks if `i++` equals 1, so it might print 2 since `i` gets incremented after the comparison.
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Dante
10 months ago
Alright, time to put my C++ knowledge to the test. I'll work through this step-by-step and make sure I understand the output.
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Martina
10 months ago
I think I've got a handle on this. The key is understanding how the post-increment operator works in the `if` condition.
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Denny
11 months ago
I'm a bit confused by this one. The `if` statement and the `cout` statements don't seem to match up at first glance.
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Karon
11 months ago
Okay, let me see here. The post-increment operator `i++` is used in the `if` condition, so I'll need to pay close attention to that.
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Amina
11 months ago
Hmm, this looks like a tricky one. I'll need to carefully step through the code and think about the order of operations.
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Basilia
1 year 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.
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Nohemi
1 year ago
User1: Got it, thanks for clarifying!
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Denny
1 year ago
User3: Oh, I see! Thanks for the explanation.
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Julio
1 year ago
User2: Actually, it prints: 2
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Jillian
1 year ago
User1: I think it prints: 1
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Regenia
1 year 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.
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Deane
1 year ago
The output is obviously 2, since the post-increment operator `i++` returns the original value of `i` before incrementing it.
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Audry
1 year ago
Correct, the output is 2.
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Audry
1 year ago
It prints: 2
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Tegan
1 year ago
I think it prints 2 because the post-increment operator is used.
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Freeman
1 year ago
I'm not sure, but I think it's A.
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Keshia
1 year ago
I believe it prints 2.
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Brock
1 year ago
I think the answer is B.
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