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C++ Institute CPA-21-02 Exam - Topic 1 Question 37 Discussion

What happens when you attempt to compile and run the following code?#include #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 {public:void set() {y = 4; z = 2;}void Print() {cout
A) It prints: 42
B) It prints: 44
C) It prints: 22
D) It prints: 2

C++ Institute CPA-21-02 Exam - Topic 1 Question 37 Discussion

Actual exam question for C++ Institute's CPA-21-02 exam
Question #: 37
Topic #: 1
[All CPA-21-02 Questions]

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

#include

#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 {

public:

void set() {

y = 4; z = 2;

}

void Print() {

cout << y << z;

}

};

int main () {

B b;

b.set();

b.Print();

return 0;

}

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Suggested Answer: A

by Van at Mar 21, 2025, 04:40 PM

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Yvonne
5 months ago
So, y gets updated but x is private? Interesting!
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Gary
5 months ago
Totally agree, that's the output!
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Colette
6 months ago
It prints: 42
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Cassi
6 months ago
I thought it would print 44.
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Yvonne
6 months ago
Wait, are you sure about that?
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Rory
6 months ago
I feel like it should print 44 because z is public and gets set to 2, but I’m not entirely confident about the initial values of y and z.
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Eric
7 months ago
I practiced a similar question, and I think the output depends on the order of initialization. I’m a bit confused about what gets printed first.
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Tyisha
7 months ago
I remember something about protected members being accessible in derived classes, so y should be set to 4 in set(). That might mean it prints 42?
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Frank
7 months ago
I think the constructor of class A initializes x, y, and z, but I'm not sure how that affects the output when we call Print in class B.
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Bonita
7 months ago
Hmm, I'm not entirely sure about this one. I think the protected member y in A is accessible to B, but I'm not 100% confident. I'll need to carefully walk through the code step-by-step to make sure I understand how the inheritance and access modifiers are working here.
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Margurite
7 months ago
I've got this! The key is understanding how the protected member y in the base class A is accessible to the derived class B. When B calls the set() method, it sets y to 4 and z to 2. Then the Print() method outputs those values, so the final output is 42.
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Huey
8 months ago
I'm a bit confused by the access modifiers here. I know protected members can be accessed by derived classes, but I'm not sure how that applies to the y and z members. I'll need to review my notes on inheritance and access control to figure this one out.
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William
8 months ago
Okay, let me think this through step-by-step. The base class A has a private member x, a protected member y, and a public member z. The derived class B inherits publicly from A and has access to y and z. The set() method in B sets y to 4 and z to 2. The Print() method then prints y and z, so I think the output should be 42.
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Sylvia
8 months ago
Hmm, this looks like a tricky one. I'll need to carefully trace the code to understand how the inheritance and access modifiers are working here.
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Alease
1 year ago
Ha! This is a classic trick question. The answer is obviously C) It prints: 22. Why, you ask? Because the protected member y is set to 4, and the public member z is set to 2. So when we print them, we get 42. Easy as pie!
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Yoko
1 year ago
No way, it prints: 22. Trust me on this one.
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Tula
1 year ago
I think you're mistaken. It actually prints: 42
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Vincenza
1 year ago
I believe it prints: 44 too, because the Print function in class B outputs the values of y and z
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Claudia
1 year ago
Hmm, this is tricky. I'm going to have to think about it carefully. Wait, I've got it! The protected member y is accessible in the derived class, so we can modify it. And since z is public, we can also change its value. The output should be 44. I'll go with B) It prints: 44.
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Yvonne
1 year ago
So, the correct answer is B) It prints: 44.
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Kina
1 year ago
Yes, and since z is public, its value can also be changed.
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Bethanie
1 year ago
I agree, the protected member y can be modified in the derived class.
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Celestine
1 year ago
I think the output will be 44.
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Avery
1 year ago
I'm not sure, but I think it prints: 44 as well
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Pearly
1 year ago
I agree with Cammy, because the set function in class B changes the value of y to 4 and z to 2
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Cammy
1 year ago
I think it prints: 44
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Shayne
1 year ago
Okay, let's think this through. The default constructor of class A initializes x to 1, y to 2, and z to 3. In the set() function, we change y to 4 and z to 2. So the output should be 42. I'll go with B) It prints: 44.
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Daryl
1 year ago
Ah, a classic inheritance question. The protected member y is accessible in the derived class B, so we can modify it in the set() function. And since z is public, we can also change its value. The output should be 44. This is an easy one!
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Melina
1 year ago
Hmm, let's see. The code creates an object of class B, which inherits from class A. The set() function sets the value of the protected member y to 4 and the public member z to 2. The Print() function then prints the values of y and z. I think the answer is B) It prints: 44.
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Yvonne
1 year ago
Actually, it prints: 42.
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Laticia
1 year ago
Actually, it prints: 44.
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Ricarda
1 year ago
No, I believe it prints: 42.
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Amber
1 year ago
No, I believe it prints: 42.
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Willodean
1 year ago
I think the answer is B) It prints: 44.
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Brynn
1 year ago
I think the answer is B) It prints: 44.
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