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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Ashley

8 days ago

Hmm, I'm not sure about this one. Let me think it through... Oh, I got it! The 'y' variable is protected, so the derived class can access it. And the 'z' variable is public, so it can be printed. C is the right answer, I think.

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Jerry

9 days 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!

upvoted 0 times

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Stephen

13 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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