Partial specialization of a method in a templated

问题:

Given:

struct A
{
    virtual bool what() = 0;
};

template<typename T, typename Q>
struct B : public A
{
    virtual bool what();
};

I want to partially specialize what like:

template<typename T, typename Q>
bool B<T, Q>::what()
{
    return true;
}

template<typename Q>
bool B<float, Q>::what()
{
    return false;
}

But it appears that this isn't possible (is it in C++11?) so I tried SFINAE:

template<typename T>
typename std::enable_if<std::is_same<T, float>::value, bool>::type B<T>::what()
{
    return true;
}

template<typename T>
typename std::enable_if<!std::is_same<T, float>::value, bool>::type B<T>::what()
{
    return false;
}

This also doesn't work, I have no idea why though, does anyone? So I found this thread and ended up with:

template<typename T, typename Q>
struct B : public A
{
    virtual bool what()
    {
        return whatimpl(std::is_same<T, float>());
    }

    bool whatimpl(std::false_type)
    {
        return false;
    }

    bool whatimpl(std::true_type)
    {
        return true;
    }
};

This final solution works, but why doesn't the enable_if technique work? I'm also very open to suggestions of a cleaner answer that I haven't encountered yet.

I simplified my examples as much as possible - in my real use case what() isn't called what and actually does a fair bit of work, and I'll want to 'specialize' on a user defined type, not float.

回答1:

Partial specialization is explicitly permitted by the standard only for class templates (see 14.5.5 Class template partial specializations)

For members of class template only explicit specialization is allowed.

14.7 (3) says:

An explicit specialization may be declared for a function template, a class template, a member of a class template or a member template. An explicit specialization declaration is introduced by template<>.

So any definition starting with

template<typename T>  

is not an allowed syntax for member of class template specialization.

[edit]

As to SFINAE attempt, it failed because actually there are neither overloads nor specializations here (SFINAE works while defining a set of candidate functions for overload resolution or while choosing proper specialization). what() is declared as a single method of class template and should have a single definition, and this definition should have a form:

template<typename T, typename Q> 
B<T,Q>:: bool what(){...}

or may be also explicitly specialized for particular instantiation of class B:

template<> 
B<SomeParticularTypeT,SomeParticularTypeTypeQ>:: bool what(){...}

Any other forms are syntacticaly invalid, so SFINAE can't help.

回答2:

Why not just change it to..

template<typename T, typename Q> 
struct B : public A 
{   
   bool what()
   {
      return false; //Or whatever the default is...
   }
}; 

template<typename Q>
struct B<float, Q> : public A 
{   
   bool what()
   {
      return true;
   }
}; 

回答3:

Two possible solutions depending on your use case:

1. Flexible implementation classes: One problem with your method is that there is a lack of flexibility in the types - no more than true or false type. Based on this excellent CppCon talk(Slide 77), I wrote up this solution by delegating the work to another implementation specific class template. You can see and run this code here. The drawback to this method is that I can't access the rest of the class members but they can be passed in.
2. Manipulating Enable If: I haven't fully understood why your enable_if solution didnt work but here is mine and it is working. It allows you to partially specialize within the class itself.

P.S. I'm trying to add the code here directly but there is some formatting problem. If someone could help me out with adding the formatted code from Coliru it'd be great.