C++ Institute CPA-21-02 Exam - Topic 7 Question 38 Discussion

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

Question #: 38
Topic #: 7

[All CPA-21-02 Questions]

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

#include

using namespace std;

class A {

int x;

protected:

int y;

public:

int z;

A() { x=1; y=2; z=3; }

};

class B : public A {

string z;

public:

void set() {

y = 4;

z = "John";

}

void Print() {

cout << y << z;

}

};

int main () {

B b;

b.set();

b.Print();

return 0;

}

AIt prints: 4John

BIt prints: 2John

CIt prints: 23

DIt prints: 43

Show Suggested Answer

Suggested Answer: A

by My at Apr 07, 2025, 03:32 AM

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Phung

8 months ago

The protected variable y is accessible in B, so 4 makes sense.

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Joanna

9 months ago

I thought the output would be 23, not 4John.

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Irma

9 months ago

Totally agree, it's definitely 4John!

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Keshia

9 months ago

Wait, how does it print 4? I thought it would be 2.

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Tamar

9 months ago

It prints: 4John

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Edmond

10 months ago

I’m confused about the `z` variable in class B. Does it overshadow the `z` in class A? I might lean towards 2John, but I’m not certain.

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Benedict

10 months ago

I practiced a similar question where protected members were accessed in derived classes. I feel like it should be 4John.

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Ronald

10 months ago

I'm not entirely sure, but I remember something about variable shadowing. Maybe it prints 2John instead?

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Luisa

10 months ago

I think the output will be 4John because the `set()` function modifies the protected variable `y` and the `z` in class B.

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Kaycee

10 months ago

I'm not entirely sure about this one. There seem to be a few things going on with the variable names and access modifiers. I'll need to work through it step-by-step to figure out the correct output.

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Jamal

11 months ago

Okay, I've got this! The key is understanding that the protected member `y` in the base class `A` can be accessed by the derived class `B`. So the output should be `4John`.

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Luther

11 months ago

Hmm, I'm a bit confused by the class hierarchy and access specifiers here. I'll need to review my notes on inheritance and think through how the derived class can access the base class members.

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Shalon

11 months ago

This looks like a question on inheritance and access modifiers. I'll need to carefully trace the code to understand how the variables and methods are being accessed across the derived class.

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Isaac

1 year ago

Wow, talk about a trick question! I'm going to guess C, just because the output seems a bit unexpected. Who knows, maybe the compiler will throw a curveball!

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Hildegarde

1 year ago

User3: Yeah, it's definitely a tricky one

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Martina

1 year ago

User2: I agree, the output does seem a bit unexpected

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Novella

1 year ago

User1: I think it prints: 23

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Elise

1 year ago

This is a tricky one, but I'm going with A. The protected `y` is accessible in `B`, and the `z` in `B` hides the `z` in `A`, so it should print 4John.

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Ramonita

1 year ago

Yes, the output will be 4John

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Serina

1 year ago

I agree, the protected variable y can be accessed in B, so it should be 4John

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Ryan

1 year ago

I think it prints: 4John

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Princess

1 year ago

I'm pretty sure it's D. The `z` member in `B` hides the `z` member in `A`, so it will print 43.

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Yun

1 year ago

That makes sense. So the correct answer is D.

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Norah

1 year ago

Yes, the output will be 43 because the z member in B hides the z member in A.

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Thaddeus

1 year ago

I think you're right. The code will print 43.

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India

1 year ago

Hmm, I think the answer is B. The protected member `y` is accessible in the derived class `B`, so it will print 4John.

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Wei

1 year ago

Exactly, and the string `z` in the derived class `B` is set to 'John' in the set() function, which is then printed along with the modified value of `y`.

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