无法访问朋友类的私人会员

时间:2013-02-28 20:01:42

标签: c++ header private friend

有人会介意帮我解决C ++链接/编码难题吗?

我有一个类Shape。 Shape需要使用类Center的私有数据成员,x和y坐标。我宣布朋友类Shape;然后#include" center.h"在Shape.h中。在Shape.cpp中,我定义了我的ostream&运营商LT;< (ostream& ostr,const Center& c)函数,它使用c.xCord; c.yCord访问中心的私人数据成员。

当我尝试编译Shape.cpp时,我得到了那些数据变量的访问错误,比如我还没有将Shape声明为友元类。我觉得这与编译时的链接顺序有关。我该如何解决这个问题?

#ifndef CENTER_H
#define CENTER_H

class Center
{
    public:
        Center(double x, double y) { xCord = x; yCord = y; }
            // constructor
        friend class Shape;
            // allow Shape to use Center's x and y values

    private:
        double xCord;
            // X-coordinate
        double yCord;
            // Y-coordinate
};

#endif

#ifndef SHAPE_H
#define SHAPE_H

#include "center.h"
#include <iostream>
using namespace std;

class Shape
{
    public:
        Shape(double x, double y) : s_center(x, y) {}
            // constructor
        void moveCenter();
            // moves the center of the shape
        friend ostream& operator<< (ostream& ostr, const Center& c);
            // allows the printing of the Center object

        virtual void printCenter();
            // returns the center of the shape
        virtual double printArea();
            // returns the area of the shape

        virtual bool checkSurface(Shape& s) = 0;
            // checks if the shape can fit into
            // a given surface
        virtual double findArea() = 0;
            // calculates the area of the shape

    private:
        Center s_center;
            // center of the shape
};

#endif

// in shape.cpp
ostream& operator<< (ostream& ostr, const Center& c)
{
    ostr << "(" << c.xCord << ", " << c.yCord << ")";
    return ostr;
}

2 个答案:

答案 0 :(得分:4)

根据C ++ 11标准的第11.3 / 10段:

  

友谊既不是继承的也不是传递。 [...]

如果班级A是班级friend的{​​{1}},而功能B是班级f()的{​​{1}},则不会同时friend A f()friend

如果您希望B访问operator <<的私有成员变量,则应将friend声明为CenterCenter

#ifndef CENTER_H
#define CENTER_H

#include <ostream>

class Center
{
public:
    Center(double x, double y) { xCord = x; yCord = y; }

    friend class Shape;
    friend std::ostream& operator<< (std::ostream& ostr, const Center& c);
//  ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

private:
    double xCord;
    double yCord;
};

#endif

答案 1 :(得分:1)

您的运营商&lt;&lt;与类形状无关。友谊不会扩展到运营商。您需要将该特定运营商声明为朋友。