c++ - C++重载后缀运算符

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What are the basic rules and idioms for operator overloading?
                                
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根据我的理解，重载后缀运算符对变量的影响将反映在变量的下一次出现中。

但是下面的程序与我的理解相矛盾，

请帮助我了解以下程序中发生的情况。

#include <iostream>

typedef struct Rectangle_Tag
{
    int len;
    int breadth;

}Rectangle_S;

/* Overloading an infix opertor */
Rectangle_S operator+(Rectangle_S a, Rectangle_S b)
{
    Rectangle_S c;

    c.len = a.len + b.len;
    c.breadth = a.breadth + b.breadth;

    return c;
}

/* Overloading a prefix opertor */
Rectangle_S operator++(Rectangle_S &a)
{
    a.len += 1;
    a.breadth += 1;

    return a;
}

/* Overloading a postfix opertor */
Rectangle_S operator++(Rectangle_S &a, int val)
{
    a.len += 1;
    a.breadth += 1;

    return a;
}

int main(void)
{
    Rectangle_S r1, r2, r3;

    r1.len = 20;
    r1.breadth = 10;

    r2.len = 20;
    r2.breadth = 10;

    r3 = (r1++) + (r2);

    std::cout << "\tr3.len     : " << r3.len << '\n';
    std::cout << "\tr3.breadth : " << r3.breadth << '\n';

    return (0);
}


//Expected Output :
    r3.len     : 40
    r3.breadth : 20

//Actual Output :
    r3.len     : 41
    r3.breadth : 21

最佳答案

没有人是完美的！

该代码的作者以一种特殊的方式实现了后缀运算符。正确的方法是

Rectangle_S operator++(Rectangle_S &a, int)
{
    auto old = a;
    ++a;
    return old;
}

还要注意，前缀++应该通过引用返回修改后的对象。