模板类中的模板专门化

Question

template <typename T>
class A
  : public std::map<std::string, double (T::*)() const>
{
  ...

  void create();

  ...
};

template <typename T>
void A<T>::create()
{ assert(false); }

template<class N1, class N2>
class C_0: public B<N1, N2>
{
  ...
}

...

template<class N1, class N2>
some_namespace::A< some_namespace::C_0<N1, N2> >
  some_namespace::C_0<N1, N2>::A_;

template <>
void some_namespace::A< C_0<N1, N2> >::create() // line 1151
{
  blah_(&C_0<N1, N2>::get_stuff);
}

C_0.h:1151:45: error: 'N1' was not declared in this scope
C_0.h:1151:49: error: 'N2' was not declared in this scope
C_0.h:1151:51: error: template argument 1 is invalid
C_0.h:1151:51: error: template argument 2 is invalid
C_0.h:1151:53: error: template argument 1 is invalid
C_0.h:1151:63: error: 'create' is not a template function

还尝试在专门化的C_0中进行模板方法专门化：

namespace some_namespace
{
  class C_0_specialized: public C_0<double_t, size_t>
  {};

  template <>
  void A<C_0_specialized>::create()
  {
    blah_(&C_0_specialized::get_stuff);
  }
}

这一次它编译，但调用原始方法，即void a ::create（）{assert（false）；}

解决方案

template<class N1, class N2>
class A< C_0<N1, N2> >
{

  ...

  void create()
  {
    blah_(&C_0<N1, N2>::get_stuff);
  }

  ...

}

Answer 1

首先，您忘记告诉编译器n1和n2是类型

template<class N1, class N2>
void some_namespace::A< C_0<N1, N2> >::create() // line 1151
{
  blah_(&C_0<N1, N2>::get_stuff);
}

这仍然不起作用，因为没有定义类A > 。首先声明。注意：您确实定义了一个 ，但这是一般情况，而不是您不试图定义成员的特殊情况。

顺便说一句，以下三行完全没有任何意义。

template<class N1, class N2>
some_namespace::A< some_namespace::C_0<N1, N2> >
  some_namespace::C_0<N1, N2>::A_;