Question

Please look at the following code:

#include <iostream>
using namespace std;

class A {
  public:
    A() {};    
    virtual void foo(double d) { cout << d << endl; }
    virtual void foo(double d, int a) = 0;
  };

class B : public A {
  public: 
    B() {};
    virtual void foo(double d, int a) { cout << d << endl << a << endl; }  
  };

int main()
  {
  B b;
  b.foo(3.14);
  return 0;
  }

The compiler (tried g++ and visual c++ 2008) says that there’s no function like B:foo(double). The exact message of g++ is:

main.cpp:21: error: no matching function for call to ‘B::foo(double)’

It looks like the effect of hiding rule, but in my opinion the rule should not be used here, since I’m not overriding foo(double) and both foo methods are defined in base class.

I know that I can fix the problem with

using A::foo;

declaration in the derived class B.

Can you explain why the code does not compile and what rules of C++ apply here?

Answer 1

The hiding rule is not about overriding, it is about hiding of names. If the derived class declares a member function, this hides other base class member functions with the same name. This also happens in your case.