有班级好友但无法访问私有成员

发布于 2024-09-10 10:35:04 字数 550 浏览 9 评论 0原文

友元函数应该能够访问类的私有成员，对吗？那么我在这里做错了什么？我已将 .h 文件包含在运算符<< 中我想和班级成为朋友。

#include <iostream>

using namespace std;
class fun
{
private:
    int a;
    int b;
    int c;


public:
    fun(int a, int b);
    void my_swap();
    int a_func();
    void print();

    friend ostream& operator<<(ostream& out, const fun& fun);
};

ostream& operator<<(ostream& out, fun& fun)
{
    out << "a= " << fun.a << ", b= " << fun.b << std::endl;

    return out;
}

原文

Friend functions should be able to access a class private members right?
So what have I done wrong here? I've included my .h file with the operator<< I intent to befriend with the class.

#include <iostream>

using namespace std;
class fun
{
private:
    int a;
    int b;
    int c;


public:
    fun(int a, int b);
    void my_swap();
    int a_func();
    void print();

    friend ostream& operator<<(ostream& out, const fun& fun);
};

ostream& operator<<(ostream& out, fun& fun)
{
    out << "a= " << fun.a << ", b= " << fun.b << std::endl;

    return out;
}

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泪痕残 2024-09-17 10:35:04

在这里......

ostream& operator<<(ostream& out, fun& fun)
{
    out << "a= " << fun.a << ", b= " << fun.b << std::endl;

    return out;
}

你需要

ostream& operator<<(ostream& out, const fun& fun)
{
    out << "a= " << fun.a << ", b= " << fun.b << std::endl;

    return out;
}

（我已经被这个问题咬了很多次了；你的运算符重载的定义与声明不太匹配，所以它被认为是一个不同的函数。）

In here...

ostream& operator<<(ostream& out, fun& fun)
{
    out << "a= " << fun.a << ", b= " << fun.b << std::endl;

    return out;
}

you need

ostream& operator<<(ostream& out, const fun& fun)
{
    out << "a= " << fun.a << ", b= " << fun.b << std::endl;

    return out;
}

(I've been bitten on the bum by this numerous times; the definition of your operator overload doesn't quite match the declaration, so it is thought to be a different function.)

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寂寞清仓 2024-09-17 10:35:04

签名不匹配。您的非会员功能很有趣& fun，朋友声明为 const fun&乐趣。

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空袭的梦i 2024-09-17 10:35:04

您可以通过在类定义中编写友元函数定义来避免此类错误：

class fun
{
    //...

    friend ostream& operator<<(ostream& out, const fun& f)
    {
        out << "a= " << f.a << ", b= " << f.b << std::endl;
        return out;
    }
};

缺点是每次对 operator<< 的调用都是内联的，这可能会导致代码膨胀。

（另请注意，该参数不能称为 fun，因为该名称已经表示一种类型。）

You can avoid these kinds of errors by writing the friend function definition inside the class definition:

class fun
{
    //...

    friend ostream& operator<<(ostream& out, const fun& f)
    {
        out << "a= " << f.a << ", b= " << f.b << std::endl;
        return out;
    }
};

The downside is that every call to operator<< is inlined, which might lead to code bloat.

(Also note that the parameter cannot be called fun because that name already denotes a type.)

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