Question

Suppose I have some code like this:

class Base {
    public:
      virtual int Foo(int) = 0;
};

class Derived : public Base {
    public:
      int Foo(int);
      virtual double Foo(double) = 0;
};

class Concrete : public Derived {
    public:          
      double Foo(double);
};

If I have a object of type Concrete, why can I not call Foo(int)?
If I change the name of Foo(double) so that it isn t overloading Foo, then all is well and both methods are accessible, but this isn t what I want.
Similarly, if I remove Concrete class and implement Foo(double) in Derived, then both are accessible, but again, not what I want.

Answer 1

Name lookup happens before overload resolution, so once Foo has been found in Concrete, base classes won t be search for other methods called Foo. int Foo(int) in Derived is hidden by the Foo in Concrete.

You have a number of options.

Change the call to be explicit.

concrete.Derived::Foo(an_int);

Add a using declaration to Concrete.

class Concrete : public Derived {
public:          
   using Derived::Foo;
   double Foo(double);
};

Call the function through a base reference.

Derived& dref = concrete;
dref.Foo(an_int);

Answer 2

Foo(double) hides the function from your base. You can make it visible though:

class Concrete : public Derived 
{
public:          
  using Derived::Foo;
  double Foo(double);
};

友情链接