Ambiguous when two superclasses have a member function with the same name, but different signatures

Question

struct A {
    void f(int x) {}
};

struct B {
    template void f(T x) {}
};

struct C : public A, public B {};

struct D {
    void f(int x){}
    te         


        

Answer 1

Consider this simpler example:

struct A{
 void f(int x){}
};

struct B{
 void f(float t){}
};


struct C:public A,public B{
};

struct D{
 void f(float n){}
 void f(int n){}
};


int main(){
 C c;
 c.f(3);

 D d;
 d.f(3);
}

In this example, same as yours, D compiles but C does not.
If a class is a derived one, member lookup mechanism behaves different. It checks each base class and merges them: In the case of C; Each base class matches the lookup ( A::f(int) and B::f(float) ). Upon merging them C decides they are ambiguous.

For the case class D: int version is selected instead of float because parameter is an integer.