explicit specialization of template class member function

问题

I need to specialize template member function for some type (let\'s say double). It works fine while class X itself is not a template class, but when I make it template GCC starts giving compile-time errors.

#include <iostream>
#include <cmath>

template <class C> class X
{
public:
   template <class T> void get_as();
};

template <class C>
void X<C>::get_as<double>()
{

}

int main()
{
   X<int> x;
   x.get_as();
}

here is the error message

source.cpp:11:27: error: template-id
  \'get_as<double>\' in declaration of primary template
source.cpp:11:6: error: prototype for
  \'void X<C>::get_as()\' does not match any in class \'X<C>\'
source.cpp:7:35: error: candidate is:
  template<class C> template<class T> void X::get_as()

How can I fix that and what is the problem here?

Thanks in advance.

回答1:

It doesn't work that way. You would need to say the following, but it is not correct

template <class C> template<>
void X<C>::get_as<double>()
{

}

Explicitly specialized members need their surrounding class templates to be explicitly specialized as well. So you need to say the following, which would only specialize the member for X<int>.

template <> template<>
void X<int>::get_as<double>()
{

}

If you want to keep the surrounding template unspecialized, you have several choices. I prefer overloads

template <class C> class X
{
   template<typename T> struct type { };

public:
   template <class T> void get_as() {
     get_as(type<T>());
   }

private:
   template<typename T> void get_as(type<T>) {

   }

   void get_as(type<double>) {

   }
};

回答2:

If one is able to used std::enable_if we could rely on SFINAE (substitution failure is not an error)

that would work like so:

#include <iostream>
#include <type_traits>

template <class C> class X
{
public:
    template <class T, typename std::enable_if< ! std::is_same<double,T>::value>::type * = nullptr > void get_as(){
        std::cout << "get as T" << std::endl;
    }


    template <class T, typename std::enable_if< std::is_same<double,T>::value>::type * = nullptr  > void get_as(){
        std::cout << "get as double" << std::endl;
    }
};


int main()
{

    X<int> d;
    d.get_as<double>();

   return 0;
}

The ugly thing is that, with all these enable_if's only one specialization needs to be available for the compiler otherwise disambiguation error will arise. Thats why the default behaviour "get as T" needs also an enable if.

来源：https://stackoverflow.com/questions/5512910/explicit-specialization-of-template-class-member-function

标签

c++

templates

gcc

specialization