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我希望有人能阐明这段代码片段，这让我感到困惑。

   //-------------------------------------------------------------------------------
   // 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)

为什么不Wrapper_To_Call_Display接受任何参数，尽管它应该char*根据它的声明接受一个参数？

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，但是为什么函数调用DoItB(TClassB::Wrapper_To_Call_Display)作为TClassB::Wrapper_To_Call_Display参数甚至很难它不是指针？

提前感谢

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

4

1 回答 1

3

在 C/C++ 中，当使用不带参数的函数名（即没有括号）时，它是指向函数的指针。TClassB::Wrapper_To_Call_Display指向实现函数代码的内存地址的指针也是如此。

因为TClassB::Wrapper_To_Call_Display是一个指向一个void函数的指针，它需要一个char*它的时间，void (*)(char* test)所以它匹配所需的类型DoItB。

于 2012-04-08T04:57:21.127 回答