C++ Institute Exam CPA Topic 6 Question 89 Discussion

Actual exam question for C++ Institute's CPA - C++ Certified Associate Programmer exam

Question #: 89
Topic #: 6

[All CPA - C++ Certified Associate Programmer Questions]

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

#include

using namespace std;

class A {

public :

void print() {

cout << "A ";

}

};

class B {

public :

void print() {

cout << "B ";

}

};

int main() {

B sc[2];

B *bc = (B*)sc;

for (int i=0; i<2;i++)

(bc++)->print();

return 0;

}

AIt prints: A A

BIt prints: B B

CIt prints: A B

DIt prints: B A

Show Suggested Answer

Suggested Answer: B

by Amie at May 29, 2024, 02:19 AM

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Eugene

2 months ago

Woah, this question really tests your understanding of object-oriented programming and pointers. I'm going to go with B, because the code is clearly working with B objects, even though it's doing some fancy pointer stuff. Gotta love these tricky certification questions!

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Providencia

23 days ago

Yeah, I agree. It's all about working with those B objects.

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Providencia

1 months ago

I think it prints: B B

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Leah

2 months ago

This is a classic pointer-related question. I'm confident the answer is B. The code is explicitly creating an array of B objects and then calling the print() method on each one, so the output should be 'B B'.

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Gerald

2 months ago

Yes, the output should be 'B B' since it's iterating through the array of B objects.

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Gwenn

2 months ago

I agree, the code is creating an array of B objects and calling the print() method on each one.

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Horace

2 months ago

That's right, the code is calling the print() method on each B object in the array.

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Iluminada

2 months ago

I think the answer is B too. It makes sense with the code logic.

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Elenor

2 months ago

Yeah, it makes sense because it's creating an array of B objects.

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Ilene

2 months ago

Haha, this question is like a magic trick! The answer must be C. The pointer is casting the array of B objects to a pointer to B objects, but the print() method being called is the one from the A class. The output will be a mix of 'A' and 'B'.

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Vanesa

2 months ago

Hmm, I'm not sure about this one. The code looks a bit tricky, but I'm leaning towards D. The way the pointer is being incremented and cast, it might end up calling the print() method of the A class instead. Let's see what the others think.

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Francine

2 months ago

I think the answer is B. The code creates an array of B objects, then casts it to a pointer to B objects and calls the print() method on each one. Since the print() method is defined in the B class, it should print 'B B'.

upvoted 0 times