Virtual function implemented in base class not bei

问题:

I've got a situation where it seems like the compiler isn't finding the base class definition/implementation of a virtual function with the same name as another member function.

struct One {};

struct Two {};

struct Base
{
    virtual void func( One & );
    virtual void func( Two & ) = 0;
};

struct Derived : public Base
{
    virtual void func( Two & );
};

void Base::func( One & )
{}

void Derived::func( Two & )
{}

// elsewhere
void this_fails_to_compile()
{
    One one;
    Derived d;
    d.func( one );
}

I'm using Visual C++ 2008. The error message is:

error C2664: 'Derived::func' : cannot convert parameter 1 from 'One' to 'Two &'

I would have thought that the type based dispatch would work and call the defined base class function. If I add a Derived::func( One & ) it does compile and get called correctly, but in my situation, that version of the function can be done in the base class and usually derived classes don't need to implement it themselves. I'm currently working around it by putting a differently named, non-virtual function in the base class that forwards the call to function causing the problem:

// not virtual, although I don't think that matters
void Base::work_around( One & one )
{
    func( one );
}

That works but is obviously less than ideal.

What inheritance and/or name-hiding rule am I missing here?

回答1:

You are hiding the method in the derived class. The simplest solution is to add a using declaration to the derived class.

struct Derived : public Base
{
    using Base::func;
    virtual void func( Two & );
};

The issue is that when the compiler tries to lookup the func identifier in the call d.func(one) it has to do that from Derived upwards, but it will stop in the first context where it finds the func identifier, which in this case it is Derived::func. No further lookup is performed and the compiler was seeing only the Derived::func( Two& ).

By adding the using Base::func; directive, when the compiler sees the Derived definition it brings all of Base::func declarations into scope, and it will find that there is a Base::func( One & ) that was not overridden in Derived.

Note also that if you were calling through a reference to Base, then the compiler would find both overloads of func and would dispatch appropriately each one to the final overrider.

Derived d;
Base & b = d;
b.func( one ); // ok even without the 'using Base::func;' directive

回答2:

You hide func(One&) function in Derived. You could use fully qualified name:

d.Base::func( one );