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C++ Institute Exam CPA Topic 7 Question 84 Discussion

Actual exam question for C++ Institute's CPA exam
Question #: 84
Topic #: 7
[All CPA Questions]

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

#include

#include

using namespace std;

class A {

public:

A() { cout << "A no parameters";}

A(string s) { cout << "A string parameter";}

A(A &a) { cout << "A object A parameter";}

};

class B : public A {

public:

B() { cout << "B no parameters";}

B(string s) { cout << "B string parameter";}

B(int s) { cout << "B int parameter";}

};

int main () {

A a2("Test");

B b1(10);

B b2(b1);

return 0;

}

Show Suggested Answer Hide Answer
Suggested Answer: B

by Alishia at Apr 10, 2024, 03:07 AM

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Freida
12 months ago
I think the answer is C as well. The increment in the pointer changes the print order.
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Glennis
12 months ago
I'm not sure, but I think it might be D because of how the pointers are being used.
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Viva
1 years ago
I agree with Tatum, the code makes use of pointer arithmetic to print A B.
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Tatum
1 years ago
I think the answer is C because it alternates between A and B.
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Crista
1 years ago
I think because we are incrementing the pointer to class B and then calling the print function
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Shenika
1 years ago
But why do you think it prints: B B?
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Pamela
1 years ago
I disagree, I believe it prints: B B
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Crista
1 years ago
I feel like it may print: A B too
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Shenika
1 years ago
I think it prints: A B
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Luann
1 years ago
You know, I'm starting to wonder if the answer is actually D) B A. Think about it - we're casting the sc array to a B* pointer, but the underlying objects are actually A objects. So when we call print() on the B* pointer, it should invoke the print() function of the A class, right? Hmm, I'm not totally sure, but that's my best guess.
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Gladys
1 years ago
Haha, I bet the exam writers are just trying to trip us up with this one. They're probably hoping we'll overthink it and second-guess ourselves. I'm just going to go with my gut and say it prints B B. What could be simpler than that?
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Shawnna
1 years ago
Hold on, I'm not so sure about that. If the objects are actually A objects, shouldn't it be printing A A? Or maybe it's going to print A B since we're accessing the objects through a B* pointer? This is really confusing, I need to think this through more carefully.
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Katie
1 years ago
I think the key here is the casting of the sc array to a B* pointer. Since we're calling print() on the B* pointer, it's going to invoke the print() function of the B class, even though the underlying objects are actually A objects. So my guess is that it's going to print B B.
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Vicky
1 years ago
Okay, let's take a closer look at this. We've got two classes, A and B, both with a print() function that prints out their respective letters. Then in the main function, we create an array of B objects and cast it to a B* pointer. Finally, we loop through and call the print() function on each element. Hmm, this is tricky.
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Belen
1 years ago
I think it will print 'A B' because we are casting the array of B objects to a B* pointer and calling the print() function on each element.
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Luther
1 years ago
C) It prints: A B
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Wayne
1 years ago
Whoa, this question looks like a real mindbender! I'm not even sure what's going to happen when I run this code. Is it going to print A A, B B, A B, or B A? My brain is already starting to hurt just thinking about it.
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