将成员函数指针作为模板类型传递

Question

我不明白为什么 clang 拒绝此代码。我从我的朋友那里得到它，并在 VisualStudio 上为他编译了它......我有很多铿锵声。

#include <utility>
#include <iostream>

template< typename Signature >
class Delegate;

template< typename Ret, typename Param >
class Delegate< Ret(Param) >
{
public:

   Ret operator()(Param&& p_param)
   {
      return m_ifunc(m_obj, std::forward< Param >(p_param));
   }

   template< typename ObjType, typename Ret(ObjType::*Method)(Param) >
   friend auto createDelegate(ObjType * const p_obj)
   {
      Delegate< Ret(Param) > del;
      del.m_obj = p_obj;
      del.m_ifunc = &ifunction< ObjType, Method >;
      return del;
   }

private:

   void * const m_obj = nullptr;
   Ret (*m_ifunc)(void*, Param&&) = nullptr;

   template< typename ObjType, typename Ret(ObjType::*Method)(Param) >
   static Ret ifunction(void * const p_obj, Param&& p_param)
   {
      ObjType * const obj = (ObjType * const) p_obj;
      return (obj->*Method)(std::forward< Param >(p_param));
   }
};

struct Test
{
   void test(int x)
   {
      std::cout << x << std::endl;
   }

};

int main()
{
   Test t;

   Delegate< void(int) > d = Delegate< void(int) >::createDelegate< Test, &Test::test >(&t);

   d(5);
}

这是我得到的错误有人明白发生了什么吗？我已经看到了这种为函数指针指定模板参数的方式，我想这对 clang 的严格性有所遗漏。

main.cpp:17:41: error: expected a qualified name after 'typename'
   template< typename ObjType, typename Ret(ObjType::*Method)(Param) >
                                        ^
main.cpp:31:41: error: expected a qualified name after 'typename'
   template< typename ObjType, typename Ret(ObjType::*Method)(Param) >
                                        ^
main.cpp:52:53: error: no member named 'createDelegate' in 'Delegate<void (int)>'
   Delegate< void(int) > d = Delegate< void(int) >::createDelegate< Test, &Test::test >(&t);
                             ~~~~~~~~~~~~~~~~~~~~~~~^
main.cpp:52:69: error: 'Test' does not refer to a value
   Delegate< void(int) > d = Delegate< void(int) >::createDelegate< Test, &Test::test >(&t);
                                                                    ^
main.cpp:39:8: note: declared here
struct Test
       ^
main.cpp:52:82: error: definition or redeclaration of 'test' not allowed inside a function
   Delegate< void(int) > d = Delegate< void(int) >::createDelegate< Test, &Test::test >(&t);
                                                                           ~~~~~~^
main.cpp:52:86: error: expected ';' at end of declaration
   Delegate< void(int) > d = Delegate< void(int) >::createDelegate< Test, &Test::test >(&t);
                                                                                     ^

Answer 1

您的代码存在几个问题，我很确定它从未以当前形式在 VisualStudio 上编译过。无论如何，我无法让它在 VS2013 上编译。以下是错误：

template< typename ObjType, typename Ret(ObjType::*Method)(Param) >
//                          ^^^^^^^^
// the member function pointer is a non-type template parameter, remove typename
friend auto createDelegate(ObjType * const p_obj)
// ^^^^
// you've declared this as a friend, but call it within main() as if it is a
// static member function, change 'friend' to 'static'


template< typename ObjType, typename Ret(ObjType::*Method)(Param) >
//                          ^^^^^^^^
// same as above, remove typename
static Ret ifunction(void * const p_obj, Param&& p_param)

m_obj 数据成员是 const 指针，但您尝试将其指向 createDelegate 中的新对象

void * const m_obj = nullptr;
//     ^^^^^
// remove the const

进行这些更改后

template< typename Signature >
class Delegate;

template< typename Ret, typename Param >
class Delegate< Ret(Param) >
{
public:

   Ret operator()(Param&& p_param)
   {
      return m_ifunc(m_obj, std::forward< Param >(p_param));
   }

   template< typename ObjType, Ret(ObjType::*Method)(Param) >
   static auto createDelegate(ObjType * const p_obj)
   {
      Delegate< Ret(Param) > del;
      del.m_obj = p_obj;
      del.m_ifunc = &ifunction< ObjType, Method >;
      return del;
   }

private:

   void * m_obj = nullptr;
   Ret (*m_ifunc)(void*, Param&&) = nullptr;

   template< typename ObjType, Ret(ObjType::*Method)(Param) >
   static Ret ifunction(void * const p_obj, Param&& p_param)
   {
      ObjType * const obj = (ObjType * const) p_obj;
      return (obj->*Method)(std::forward< Param >(p_param));
   }
};

现在代码编译并输出5。 Live demo

---

要在 VS2013（发布版本，不知道 CTP）上编译，需要进行额外的更改。由于VS2013没有实现C++14对普通函数的返回类型推导，createDelegate需要明确指定返回类型

template< typename ObjType, Ret(ObjType::*Method)(Param) >
static Delegate< Ret(Param) > createDelegate(ObjType * const p_obj)
{ /* ... */ }

---

最后，只是为了确保您了解替代方案：

Test t;
auto d = std::bind(&Test::test, t, std::placeholders::_1);
d(5);    // prints 5
std::function<void(int)> d2 = std::bind(&Test::test, t, std::placeholders::_1);
d2(5);   // prints 5