c++ friend class as argument for ctor

Question

I am wondering how come this code correct and how the compiler knows to first create an obj of class A, I would think it shouldn't compile since B's ctor request an argument of type A not int

    class A 
{
    int a1;
public:
    A(int i) { cout << i << "A"<<endl;}
    friend class B;
};

class B
{
public:
    B(A a) {cout <<"B" <<a.a1;}
};

void main() 
{
    B b(7);
}

output: 7A B7

Add explicit to A(int) constructor to prevent this behavior. — hynner, Feb 08 '15 at 09:55

score 6 · Accepted Answer · answered Feb 08 '15 at 09:58

6

This has nothing to do with friend.

Because you haven't marked the A constructor explicit, an implicit conversion from int to A occurs.

answered Feb 08 '15 at 09:58

Emil Laine

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Ed Heal · Answer 2 · 2015-02-08T10:06:28.633

2

The compiler is smart enough to think Got an int .... need and A - Wait here is the constructor. Use that

edited Feb 08 '15 at 10:06

answered Feb 08 '15 at 09:55

Ed Heal

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_"The friend bit is not required."_ It actually is: `B(A a) {cout <<"B" < – πάντα ῥεῖ Feb 08 '15 at 09:56
1

My fault - sunday morning. forgot defaults to private – Ed Heal Feb 08 '15 at 09:57

score 1 · Answer 3 · answered Feb 08 '15 at 09:54

1

As far as I know the compiler calls the constructor A implicitly (because A(int)), so your code is equivalent to:

B b(A(7));

answered Feb 08 '15 at 09:54

Marco

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c++ friend class as argument for ctor

3 Answers3