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What happens when you attempt to compile and run the following code?
#include
#include
using namespace std;
class Second;
class Base {
int age;
public:
Base () { age=5; };
friend void set(Base &ob, Second &so);
void Print() { cout << age;}
};
class Second {
string name;
public:
friend void set(Base &ob, Second &so);
void Print() { cout << name;}
};
void set(Base &ob, Second &so) {
ob.age = 0; so.name = "Bill";
}
int main () {
Base a;
Second b;
set(a,b);
a.Print();
b.Print();
return 0;
}
Suggested Answer: A
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30 comments · newest first
MA
Malissa
6 months ago
Definitely 0Bill. The order of Print matters.
PE
Pearlie
6 months ago
No, it should compile fine. The friend function works.
ER
Ernest
6 months ago
I feel like there might be a compilation error.
AD
Adolph
6 months ago
Yeah, the set function modifies both objects.
ER
Erick
7 months ago
I think it prints 0Bill.
NO
Nobuko
7 months ago
No way, it works perfectly fine!
KR
Krissy
7 months ago
I think it should give a compilation error.
OZ
Ozell
8 months ago
Wait, how does it print in that order?
HO
Hollis
8 months ago
Totally agree, that's the output!
CE
Cecily
8 months ago
A) It prints: 0Bill - Yep, that's the one! The friend function gives the set() function access to the private members of both classes, so it can modify them directly.
WA
Walton
8 months ago
Hmm, I'm not sure about this one. The friend function is a bit confusing, but I think A is the right answer. Let me double-check the code just to be sure.
GW
Gwen
8 months ago
Haha, this is a classic trick question! I bet a lot of people will get tripped up by the friend function and try to say it's a compilation error.
CA
Carmelina
8 months ago
I think the correct answer is A as well. The set() function directly modifies the private members of both classes, so the output should be 0Bill.
QU
Queen
9 months ago
I’m confused about the friend function. Does it really allow access to private members? I hope it doesn’t lead to a compilation error.
CA
Carisa
9 months ago
I practiced a similar question, and I feel like the output should be "Bill0" instead.
LE
Leatha
9 months ago
I think it should compile without errors since the friend function is declared in both classes.
MA
Marti
9 months ago
This seems straightforward enough. I'll work through it step-by-step and see if I can arrive at the correct answer.
DO
Donette
9 months ago
I'm a bit confused by the order of the `Print()` calls in `main()`. I'll need to double-check the expected output.
LE
Leonor
10 months ago
Okay, I think I've got it. The friend function `set()` is modifying the private member `age` of the `Base` class, so the output should be `0Bill`.
ER
Ernie
10 months ago
It prints: 0Bill
HU
Hubert
10 months ago
The correct answer is A) It prints: 0Bill. The set() function modifies the age member of the Base object and the name member of the Second object, so the output will be 0Bill.
GR
Gregg
10 months ago
I agree, but what if there's a typo?
BU
Bulah
11 months ago
I'm not entirely sure, but I remember something about the order of printing. Could it be that it prints "0Bill"?
YO
Yolande
11 months ago
Hmm, the use of friend functions is throwing me off a bit. I'll need to review how those work with class member access.
CE
Ceola
11 months ago
This looks like a tricky one. I'll need to carefully trace the code execution to figure out what's happening.
LA
Lashon
5 months ago
So, it's definitely A) It prints: 0Bill.
MA
Marion
5 months ago
Right! First a.Print(), then b.Print().
LA
Lavina
6 months ago
But the order of Print matters too!
AS
Ashlee
6 months ago
Yeah, the set function modifies both objects.
JA
Jame
10 months ago
I think it prints 0Bill.