c++ - 使用函数 Object tr1::function<> 调用仿函数

Question

我尝试实现 Scott Mayer 书中的代码示例，该示例是关于通过函数对象调用仿函数

头文件gameCharachter.h

#ifndef GAMECHARACTER_H
#define GAMECHARACTER_H

#include <iostream>
#include <typeinfo>
using namespace std;
#include <tr1/functional>

class GameCharacter;

int defaultHealthCalc(const GameCharacter& gc);

class GameCharacter
{

public:

    typedef std::tr1::function<int (const GameCharacter&)> HealthCalcFunc;

    explicit GameCharacter(HealthCalcFunc hcf = defaultHealthCalc)

        : healthFunc(hcf)
    {
    }

    ~GameCharacter()
    {
    }

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

    HealthCalcFunc healthFunc;

};

class EyeCandyCharacter:   public GameCharacter    // another character
{

public:

    explicit EyeCandyCharacter(HealthCalcFunc hcf = defaultHealthCalc)

        : GameCharacter(hcf)
    {
        cout<<typeid(*this).name()<<"::"<<__FUNCTION__<<""<<endl;

    }                                           

};   

struct HealthCalculator                          
{
    /*explicit*/ HealthCalculator()
    {

    }

    int operator()(const GameCharacter& gc) const     // calculation function
    {
        cout<<typeid(*this).name()<<"::"<<__FUNCTION__<<""<<endl;
           return 0;
    }                                        

};

#endif // GAMECHARACTER_H

main.cpp 是:

#include "gamecharacter.h"

int main()
{
    EyeCandyCharacter ecc1(HealthCalculator());       
    ecc1.healthValue();
}

为什么 function<> 对象拒绝调用 healthvalue() 中的 operator() 函数

Answer 1

EyeCandyCharacter ecc1(HealthCalculator());

声明一个名为 ecc1 的函数，该函数采用“指向不带参数的函数并返回一个 HealthCalculator”类型的参数，并返回一个 EyeCandyCharacter。我认为这不是您的意图。