Alternate argument lists of pure virtual won't inherit (needed "using")

Question

I've got a base class with a few pure virtuals, and I want it to provide default implementations of several methods which can be overridden if needed. I've boiled my problem down to the following:

class A {};
class B {};

class Base {
public:
  virtual void foo (const A&) {};
  virtual void foo (const B&) = 0;
};

class Derived : public Base {
public:
  virtual void foo(const B&) {};
};

int main(int argv, char** argc) {
  Derived d;
  d.foo(A());
  return 0;
}

For reasons I don't understand at all, this fails on g++ 4.8.2:

g++ scratch.cpp -o scratch
scratch.cpp: In function ‘int main(int, char**)’:
scratch.cpp:17:12: error: no matching function for call to ‘Derived::foo(A)’
   d.foo(A());
        ^
scratch.cpp:17:12: note: candidate is:
scratch.cpp:12:16: note: virtual void Derived::foo(const B&)
   virtual void foo(const B&) {};
            ^
scratch.cpp:12:16: note:   no known conversion for argument 1 from ‘A’ to ‘const B&’ 

If I define foo(const B&) in Base and leave Derived empty, the problem goes away. I've read the docs and some of the questions here, but nothing has suggested why this would be a problem. In my real code, Base::foo(const A&) will build a special object b of type B from the A object, and then call foo(b).

Answer 1

Add this line to your Derived class definition:

using Base::foo;

This will solve the problem that your foo(B) definition in Derived hides both foo(A) and foo(B) functions in Base.