How to provide a explicit specialization to only one method in a C++ template class?

Question

I have a template class that looks something like this:

template<class T> class C
{
    void A();
    void B();

    // Other stuff
};

template<class T> void C<T>::A() { /* something */ }
template<class T> void C<T>::B() { /* something */ }

What I want is to provide an explicit specialization for only A while retaining the default for B and the "other stuff".

What I have tried so far is

class D { };
template<> void C<D>::A() { /*...*/ } // Gives a link error: multiple definition

Every other variant I've attempted fails with parse errors.

---

What I did:

The original problem was that the explicit specialization was in a header file so it was getting dumped into several object files and messing up the link (Why doesn't the linker notice all the instances of the symbol are the same a just shut up?)

The solution ends up being to move the explicit specialization from the header file to a code file. However to make the other users of the header file not instance the default version, I needed to place a prototype back in the header. Then to get GCC to actually generate the explicit specialization, I needed to place a dummy variable of the correct type in the code file.

Answer 1

Try

template<> inline void c<int>::A() { ... }
//         ^^^^^^

As you have defined it in a header file. Each source file that sees it will build an explicit version of it. This is resulting in your linking errors. So jsut declare it as inline.

Answer 2

Alternatively to Martin York's inline solution you could also do in your header file:

class D { };
template<> void C<D>::A(); // Don't implement here!

And supply a .cpp file with the implementation:

template<> void C<D>::A() { /* do code here */ }

So you avoid the multiple definitions by supplying a single one. This is also good to hide implementations for specific Types away from the template header file when publishing the library.