explicit specialization of template class member function

Question

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.

Answer 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>) {     } };