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将void作为参数替换为模板化方法

[英]Substitution of void as parameter to templated method

In my code I have a class which registers method's of other classes: 在我的代码中,我有一个类,它注册其他类的方法:

#include <iostream>
using namespace std;

template< typename C>
  class Reg {
  public:
    template< typename R, typename A>
      void register_f( string name, R ( C:: *method_p ) ( A)) { /*registration process*/ }

//    template< typename R>
//      void register_void_f( string name, R ( C:: *method_p ) ( void)) { /*registration process*/ }
  };

  class A {
  public:
      int f( void) { return 1; }
      void g( int x) { /* some code*/ }
  };


  int main() {
      Reg< A> r;

      r.register_f( "g", &A::g);
/*1*///  r.register_f( "f", &A::f);
/*2*///  r.register_f< int, void>( "f", &A::f);
/*3*///  r.register_void_f< int>( "f", &A::f);

      return 0;
  }

http://ideone.com/X8PNLC http://ideone.com/X8PNLC

Uncommenting line /* 2 */ gives me an error: 取消注释行/ * 2 * /给我一个错误:

template argument deduction/substitution failed: 模板参数扣除/替换失败:

In substitution of 'template void register_f(std::string, R (C::*)(A)) [with R = R; 代替'template void register_f(std :: string,R(C :: *)(A))[用R = R; A = A; A = A; C = A] [with R = int; C = A] [R = int; A = void]': A = void]':

error: invalid parameter type 'void' 错误:参数类型无效'void'

Line /* 1 / is the same as / 2 */, but with not so informative error message. 行/ * 1 /与/ 2 * / 相同 ,但没有提供如此丰富的错误消息。

I understand that to fix the problem I can use method register_void_f , but I don't want to do it because register_f is a part of the my final API. 我理解为了解决问题,我可以使用方法register_void_f ,但我不想这样做,因为register_f是我最终API的一部分。

Question> How to fix the compilation errors without introducing register_void_f ? 问题>如何在不引入register_void_f的情况下修复编译错误?

I have an idea to solve it with partial specialized register_f , but I don't know how to do it since in C++ you can't partially specialize templated method. 我有一个想法,用部分专门的register_f来解决它,但我不知道怎么做,因为在C ++中你不能部分地专门化模板化方法。

PS> I can't use C++11. PS>我不能使用C ++ 11。

Don't use void for no arguments, use nothing - something like this: 不要使用void作为参数,不使用任何东西 - 像这样:

template< typename R>
     void register_void_f( string name, R ( C:: *method_p ) ()) { /*registration process*/ }

Overload your function: 超载你的功能:

void foo( int ) {}
double bar() { return 3.14; }

template< class R, class A >
void test(  R ( *method_p ) (A)) {  }
template< class R >
void test(  R ( *method_p ) ()) {  }

int main(){
  test(foo);
  test(bar);
}

live example 实例

Converting this to it being methods should be easy. 将其转换为方法应该很容易。

You may use the following: 您可以使用以下内容:

template< typename C> class Reg;

template <typename C, typename F> struct helper;

template <typename C, typename R, typename A>
struct helper<C, R (C::*)(A)>
{
    void operator() (Reg<C>& reg, const std::string& name, R (C::*method)(A)) const { /* Your implementation */}
};

template <typename C, typename R>
struct helper<C, R (C::*)()>
{
    void operator() (Reg<C>& reg, const std::string& name, R (C::*method)()) const { /* Your implementation */}
};


template< typename C>
  class Reg {
  public:
    template< typename F>
      void register_f(const std::string& name, F method) { helper<C, F>()(*this, name, method);  /*registration process*/ }

  };

And use it that way: 并以这种方式使用它:

Reg< A> r;

r.register_f( "g", &A::g);
r.register_f( "f", &A::f);
r.register_f<int (A::*)(void)>( "f", &A::f);

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