c++ - 如何理解这个声明中的typedef

Question

最近看了Effective C++这本书，第35条里面有一个关于typedef的声明让我很困惑。

class GameCharacter; // Question1: Why use forward declaration?

int defaultHealthCalc(const GameCharacter& gc);

class GameCharacter{

    public:
        typedef int (*HealthCalcFunc)(const GameCharacter&); // Question 2: What does this mean?

        explicit GameCharacter(HealthCalcFunc hcf = defaultHealthCalc)
            : healthFunc(hcf)
        {}

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

    private:
        HealthCalcFunc healthFunc;
};

所以我的第一个问题是:为什么作者在这里使用前向声明？有什么具体原因吗？

我的第二个问题是:如何理解 typedef 声明，以及如何使用它？我只知道 typedef int MyInt;

Answer 1

So my first question is that why the author use forward declaration here?

因为函数defaultHealthCalc的声明使用了const GameCharacter&作为参数的类型，所以需要提前声明GameCharacter。

And my second question is how to understand the typedef declaration

它声明了一个类型名称HealthCalcFunc，这是一种指向函数的指针类型，它以const GameCharacter&为参数并返回int。

and how to use it?

正如代码示例所示，

explicit GameCharacter(HealthCalcFunc hcf = defaultHealthCalc) // declare a parameter with type HealthCalcFunc; 
                                                               // the default argument is function pointer to defaultHealthCalc
: healthFunc(hcf)                                              // initialize healthFunc with hcf
{}

int healthValue() const
{return healthFunc(*this); }                                   // invoke the function pointed by healthFunc

HealthCalcFunc healthFunc;                                     // declare a member with type HealthCalcFunc