C++ Institute Exam CPA-21-02 Topic 2 Question 36 Discussion

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

Question #: 36
Topic #: 2

[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;

};

class B : public A {

string name;

public:

void set() {

y = 2;

z = 3;

}

void Print() { cout << y << z; }

};

int main () {

B b;

b.set();

b.Print();

return 0;

}

AIt prints: 123

BIt prints: 000

CIt prints: 23

DIt prints: 12

Show Suggested Answer

Suggested Answer: D

by Reena at Mar 04, 2025, 03:59 AM

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Jerry

33 minutes ago

Haha, looks like someone forgot to initialize the variables. I bet the code will just print a bunch of zeros, right? Option B for the win!

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Stephen

4 days ago

The correct answer is C. The code will print 23 because the 'y' variable is protected and can be accessed by the derived class 'B', and the 'z' variable is public and can be accessed by any class.

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Bernadine

5 days ago

So the correct answer is C) It prints: 23. Got it!

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Adria

8 days ago

I agree with Alesia. The Print function will then output the values of y and z, which are 2 and 3 respectively.

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Alesia

9 days ago

I think it prints: 23 because the set function initializes y to 2 and z to 3.

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