C++中使用friend class结合parameterized constructor计算两点距离报错问题

我需要计算两点间的距离，设计了point类：

• 带参构造函数point(int a, int b)给私有成员x、y赋值
• displayPoint()成员函数打印坐标格式(x,y)
• 将dist类声明为友元类，打算让dist类通过接收两个point对象的构造函数计算并输出距离，但代码报错。

错误代码如下：

// Parameterized Constructor using Friend Class Example :-
#include <iostream>
#include <cmath>
using namespace std;

class point
{
    int x, y;
    friend class dist;

public:
    point(int a, int b)
    {
        x = a;
        y = b;
    }
    void displayPoint()
    {
        cout << "The Point is : (" << x << "," << y << ")" << endl;
    }
};
class dist  // shows error here
{
public:
    void dist(point p1, point p2)
    {
        int x_diff = (p2.x - p1.x);
        int y_diff = (p2.y - p1.y);
        int diff = sqrt(pow(x_diff, 2) + pow(y_diff, 2));
        cout << "The difference is : " << diff << endl;
    }
};
int main()
{
    point p(1, 2);
    point q(4, 6);

    point c(1, 1);
    point d(1, 1);

    point e(1, 0);
    point f(70, 0);

    dist(p, q);
    dist(c, d);
    dist(e, f);
    return 0;
}

我尝试移除dist类，把dist()改成友元函数后代码能正常运行，可行代码如下：

void dist(point p1, point p2)
{
    int x_diff = (p2.x - p1.x);
    int y_diff = (p2.y - p1.y);
    int diff = sqrt(pow(x_diff, 2) + pow(y_diff, 2));
    cout << "The difference is : " << diff << endl;
}

错误原因分析 #

• 构造函数语法错误：构造函数不能有返回值类型，你给dist类的构造函数加了void修饰，违反了C++语法规则——构造函数必须与类同名且无返回值（包括void）。
• 构造函数调用方式错误：main中dist(p, q)的写法不合法，构造函数是用来创建类对象的，正确写法应为dist obj(p, q);，且用构造函数执行计算输出不符合代码职责分离原则。
• 精度丢失问题：sqrt()和pow()函数返回double类型，用int类型的diff接收会截断小数部分，导致距离计算精度丢失。

修正方案1：修复友元类的写法 #

如果坚持用友元类实现，修正后的代码如下：

#include <iostream>
#include <cmath>
using namespace std;

class point
{
    int x, y;
    friend class dist;

public:
    point(int a, int b) : x(a), y(b) {}
    void displayPoint()
    {
        cout << "The Point is : (" << x << "," << y << ")" << endl;
    }
};

class dist
{
public:
    // 构造函数无返回值，直接执行计算逻辑
    dist(point p1, point p2)
    {
        int x_diff = p2.x - p1.x;
        int y_diff = p2.y - p1.y;
        // 用double存储避免精度丢失
        double diff = sqrt(pow(x_diff, 2) + pow(y_diff, 2));
        cout << "The distance is : " << diff << endl;
    }
};

int main()
{
    point p(1, 2);
    point q(4, 6);
    point c(1, 1);
    point d(1, 1);
    point e(1, 0);
    point f(70, 0);

    // 通过创建dist对象触发构造函数计算
    dist obj1(p, q);
    dist obj2(c, d);
    dist obj3(e, f);
    return 0;
}

修正方案2：使用友元函数（更简洁） #

友元函数的方式更符合场景需求——仅需计算距离，不需要额外维护类对象，修正精度问题后的代码如下：

#include <iostream>
#include <cmath>
using namespace std;

class point
{
    int x, y;
    // 声明友元函数
    friend void dist(point p1, point p2);

public:
    point(int a, int b) : x(a), y(b) {}
    void displayPoint()
    {
        cout << "The Point is : (" << x << "," << y << ")" << endl;
    }
};

// 实现友元函数，用double存储距离
void dist(point p1, point p2)
{
    int x_diff = p2.x - p1.x;
    int y_diff = p2.y - p1.y;
    double diff = sqrt(pow(x_diff, 2) + pow(y_diff, 2));
    cout << "The distance is : " << diff << endl;
}

int main()
{
    point p(1, 2);
    point q(4, 6);
    point c(1, 1);
    point d(1, 1);
    point e(1, 0);
    point f(70, 0);

    dist(p, q);
    dist(c, d);
    dist(e, f);
    return 0;
}

内容的提问来源于stack exchange，提问作者Aslam Sha