c++ - 回调到非静态 C++ 成员函数

Question

我希望有人能阐明这个让我感到困惑的代码片段。

   //-------------------------------------------------------------------------------
   // 3.5 Example B: Callback to member function using a global variable
   // Task: The function 'DoItB' does something that implies a callback to
   //       the member function 'Display'. Therefore the wrapper-function
   //       'Wrapper_To_Call_Display is used.

   #include <iostream.h>   // due to:   cout

   void* pt2Object;        // global variable which points to an arbitrary object

   class TClassB
   {
   public:

      void Display(const char* text) { cout << text << endl; };
      static void Wrapper_To_Call_Display(char* text);

      /* more of TClassB */
   };


   // static wrapper-function to be able to callback the member function Display()
   void TClassB::Wrapper_To_Call_Display(char* string)
   {
       // explicitly cast global variable <pt2Object> to a pointer to TClassB
       // warning: <pt2Object> MUST point to an appropriate object!
       TClassB* mySelf = (TClassB*) pt2Object;

       // call member
       mySelf->Display(string);
   }


   // function does something that implies a callback
   // note: of course this function can also be a member function
   void DoItB(void (*pt2Function)(char* text))
   {
      /* do something */

      pt2Function("hi, i'm calling back using a global ;-)");   // make callback
   }


   // execute example code
   void Callback_Using_Global()
   {
      // 1. instantiate object of TClassB
      TClassB objB;


      // 2. assign global variable which is used in the static wrapper function
      // important: never forget to do this!!
      pt2Object = (void*) &objB;


      // 3. call 'DoItB' for <objB>
      DoItB(TClassB::Wrapper_To_Call_Display);
   }

问题1:关于这个函数调用:

DoItB(TClassB::Wrapper_To_Call_Display)

为什么 Wrapper_To_Call_Display 不接受任何参数，尽管根据其声明它应该接受 char* 参数？

问题2: DoItB声明为

void DoItB(void (*pt2Function)(char* text))

到目前为止我所理解的是 DoItB 将函数指针作为参数，但为什么函数调用 DoItB(TClassB::Wrapper_To_Call_Display) 将 TClassB::Wrapper_To_Call_Display 作为论点甚至强硬它不是一个指针？

提前致谢

代码片段来源:http://www.newty.de/fpt/callback.html

Answer 1

在 C/C++ 中，当使用不带参数的函数名称时(即没有括号)，它是指向函数的指针。所以 TClassB::Wrapper_To_Call_Display 是指向内存中实现函数代码的地址的指针。

因为 TClassB::Wrapper_To_Call_Display 是一个指向 void 函数的指针，该函数采用单个 char*，所以是 void ( *)(char* test) 所以它匹配 DoItB 要求的类型。