C++ Institute CPA-21-02 Exam - Topic 6 Question 53 Discussion

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

Question #: 53
Topic #: 6

[All CPA-21-02 Questions]

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

#include

using namespace std;

void fun(int &i);

int main()

{

int i=2;

fun(i);

cout<

return 0;

}

void fun(int &i)

{

i+=2;

}

AIt prints: 2

BIt prints: 0

CIt prints: 4

DIt prints: 16

Show Suggested Answer

Suggested Answer: C

by Darci at May 02, 2026, 01:50 AM

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Gilma

5 days ago

I agree, the reference changes the value.

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Kenny

10 days ago

Yeah, because fun modifies i directly.

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Caitlin

15 days ago

I think it prints 4.

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Dierdre

20 days ago

I thought it would print 0 for some reason!

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Rory

25 days ago

Definitely C, the reference changes the original.

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Pamella

1 month ago

Wait, are you sure it doesn't print 2?

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Yuki

1 month ago

Totally agree, it modifies the value!

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Mitzie

1 month ago

It prints: 4

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Alverta

2 months ago

I’m confused about the output. I thought it might print 0, but now I’m leaning towards 4 since the function adds 2 to `i`.

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Elroy

3 months ago

I practiced a similar question where passing by reference changed the variable's value. I believe it should print 4.

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Quentin

3 months ago

I'm not entirely sure, but I remember something about references in C++. Could it be that it prints 2 instead?

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Mariann

3 months ago

I think the function modifies the value of `i` since it's passed by reference, so it should print 4.

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