如何键入一个指向方法的指针,该方法返回一个指针方法？

Question

基本上我有以下课程:

class StateMachine {
...
StateMethod stateA();
StateMethod stateB();
...
};

stateA()和stateB()方法应该能够返回指向stateA()和stateB()的指针.如何输入定义StateMethod？

Answer 1

GotW#57表示为此目的使用具有隐式转换的代理类.

struct StateMethod;
typedef StateMethod (StateMachine:: *FuncPtr)(); 
struct StateMethod
{
  StateMethod( FuncPtr pp ) : p( pp ) { }
  operator FuncPtr() { return p; }
  FuncPtr p;
};

class StateMachine {
  StateMethod stateA();
  StateMethod stateB();
};

int main()
{
  StateMachine *fsm = new StateMachine();
  FuncPtr a = fsm->stateA();  // natural usage syntax
  return 0;
}    

StateMethod StateMachine::stateA
{
  return stateA; // natural return syntax
}

StateMethod StateMachine::stateB
{
  return stateB;
}

该解决方案有三个主要优势:

1. 它根据需要解决了问题.更好的是,它是类型安全和便携式的.

2. 它的机制是透明的:你得到调用者/用户的自然语法,以及函数自己的"return stateA"的自然语法.声明.

3. 它可能没有任何开销:在现代编译器中,代理类及其存储和功能应该内联并优化为零.

Answer 2

仅使用typedef:

class StateMachine {  

 public:  

  class StateMethod;     
  typedef StateMethod (StateMachine::*statemethod)();   

  class StateMethod {  

    statemethod   method; 
    StateMachine& obj; 

   public:  

    StateMethod(statemethod method_, StateMachine *obj_)  
      : method(method_), obj(*obj_) {} 

    StateMethod operator()() { return (obj.*(method))(); }  
  };  

  StateMethod stateA()  { return StateMethod(&StateMachine::stateA, this); }  

  StateMethod stateB()  { return StateMethod(&StateMachine::stateB, this); }  

};