C ++中的回调函数

Question

在C ++中，何时以及如何使用回调函数？编辑：我想看一个简单的例子来编写一个回调函数

Answer 1

还有执行回调的C方法：函数指针

//Define a type for the callback signature,

//it is not necessary, but makes life easier

//Function pointer called CallbackType that takes a float

//and returns an int

typedef int (*CallbackType)(float);  

void DoWork(CallbackType callback)

{

  float variable = 0.0f;

  //Do calculations

  //Call the callback with the variable, and retrieve the

  //result

  int result = callback(variable);

  //Do something with the result

}

int SomeCallback(float variable)

{

  int result;

  //Interpret variable

  return result;

}

int main(int argc, char ** argv)

{

  //Pass in SomeCallback to the DoWork

  DoWork(&SomeCallback);

}

现在，如果您希望将类方法作为回调传递，则这些函数指针的声明具有更复杂的声明，例如：

//Declaration:

typedef int (ClassName::*CallbackType)(float);

//This method performs work using an object instance

void DoWorkObject(CallbackType callback)

{

  //Class instance to invoke it through

  ClassName objectInstance;

  //Invocation

  int result = (objectInstance.*callback)(1.0f);

}

//This method performs work using an object pointer

void DoWorkPointer(CallbackType callback)

{

  //Class pointer to invoke it through

  ClassName * pointerInstance;

  //Invocation

  int result = (pointerInstance->*callback)(1.0f);

}

int main(int argc, char ** argv)

{

  //Pass in SomeCallback to the DoWork

  DoWorkObject(&ClassName::Method);

  DoWorkPointer(&ClassName::Method);

}

Answer 2

Scott Meyers举了一个很好的例子：

class GameCharacter;

int defaultHealthCalc(const GameCharacter& gc);

class GameCharacter

{

public:

  typedef std::function<int (const GameCharacter&)> HealthCalcFunc;

  explicit GameCharacter(HealthCalcFunc hcf = defaultHealthCalc)

  : healthFunc(hcf)

  { }

  int healthValue() const { return healthFunc(*this); }

private:

  HealthCalcFunc healthFunc;

};

我认为这个例子说明了一切。

std::function<> 是编写C ++回调的“现代”方法。