Friend function and method with the same name

manpreet Best Answer 2 years ago

The following classes definitions declare a friend function providing an inline definition for it. I was trying to invoke the friend function from a class method with the same name of the friend function, but in order to make it work I have to access it from the enclosing namespace (which also requires a forward declaration, class C below). Why the name lookup works for class A and it doesn't work in class B? Note that the parameters of B::swap are different from those of its friend function.

#include 

struct A {
    A(int x) : v{ x } {}
    com/tag/friend">friend void swap(A& x, A& y) { std::swap(x.v, y.v); }
    void swapm(A& other) { swap(*this, other); }
private:
    int v;
};

struct B {
    B(int x) : v{ x } {}
    com/tag/friend">friend void swap(B& x, B& y) { std::swap(x.v, y.v); }
    void swap(B& other) { swap(*this, other); } // <-- This doesn't compile
private:
    int v;
};

struct C;
void swap(C& x, C& y);
struct C {
    C(int x) : v{ x } {}
    com/tag/friend">friend void swap(C& x, C& y) { std::swap(x.v, y.v); }
    void swap(C& other) { ::swap(*this, other); }
private:
    int v;
};

int main()
{
    A a1{ 1 }, a2{ 2 }; swap(a1, a2); a1.swapm(a2);
    B b1{ 3 }, b2{ 4 }; swap(b1, b2); b1.swap(b2);
    C c1{ 5 }, c2{ 6 }; swap(c1, c2); c1.swap(c2);
}

manpreet 2 years ago

inline friend? Why not static in this case? And not friend? And then create a global one that calls the static. For me, this is a bad design more than an actual problem.

The method swap should be the one doing the work instead of the friend (because no need for a friend anymore):

struct C {
    C(int x) : v{ x } {}
    void swap(C& other) { std::swap(this->v, other.v); }
private:
    int v;
};

void swap(C& x, C& y)
{
  x.swap(y);    
}