Specializing a base class' member function in a derived class template

Question

Have a look at this code:

struct foo {
    virtual int bleh() {
        return 42;
    }
};


template<typename T>
struct bar : public foo {

};

// ERROR
template<>
int bar<char>::bleh() {
    return 12;
}

I'm trying to provide a definition of base::bleh only for bar<char> , but the compiler(gcc 4.7.2) rejects my code with the following diagnostic:

template-id ‘bleh<>’ for ‘int bar<char>::bleh()’ does not match any template declaration

It seems like base::bleh is somehow hidden in bar . I've fixed this using the following definition in bar :

template<typename T>
struct bar : public foo {
    // doesn't work
    //using foo::bleh; 

    // this works
    int bleh() {
        return foo::bleh();
    }
};

But I'm curious as of why this fails to compile. Why is the compiler rejecting my code?

Answer 1

In your non-compiling example, you're attempting to specialize and define a function that hasn't been declared in the template definition of bar ... In your later example you have actually declared as well as defined the non-specialized version of the function inside the template definition of bar , which is why it compiles. From what I can tell, here is the associated language in the standard concerning why the first version won't compile (14.7.3/4):

A member function, a member function template, a member class, a member enumeration, a member class template, or a static data member of a class template may be explicitly specialized for a class specialization that is implicitly instantiated; in this case, the definition of the class template shall precede the explicit specialization for the member of the class template