重载++后缀运算符 [英] Overload the ++ postfix operator

问题描述

#include <iostream>
using namespace std;

class Time {
private:
	int hours;             // 0 to 23
	int minutes;           // 0 to 59
public:
	// required constructors
	Time(){
		hours = 0;
		minutes = 0;
	}
	Time(int h, int m){
		hours = h;
		minutes = m;
	}
	// method to display time
	void displayTime() {
		cout << "H: " << hours << " M:" << minutes << endl;
	}

	Time& operator++()
	{
		++minutes;
		if (minutes>=60)
		{
			++hours;
			minutes = minutes - 60;
		}
		return *this;
	}

	Time& Time::operator++(int){
		
		Time temp(hours, minutes);
		minutes++;
		if (minutes >= 60)
		{
			hours++;
			minutes = minutes - 60;
		}
		return *this;
	}
	
};

int main() {
	Time T1(11, 59), T2(10, 40);

	T2 = ++(++T1); // 
	T1.displayTime();        // display 12,01 ok
	T2.displayTime();        // display 12,01 ok

	T2 = (T1++)++;           
	T1.displayTime();        // display 12,03 οκ
	T2.displayTime();        // display 12,03 why?? i think it must be 12,01

	return 0;
}

我的尝试：



问题出在对象T2

What I have tried:

The problem is at the object T2

推荐答案

你的代码是正确的：

It is correct by your code:

T2 = ++(++T1);

1.你在括号中增加T1

2.你增加大括号外的T1

3.你将 T1值分配给T2



提示：使用调试器步骤进入此代码行的三个函数调用。

1. you increment the T1 in the braces
2. you increment the T1 outside the braces
3. you assign the T1 values to T2

tip: use a debugger to step into the three function calls of this code line.

后缀运算符的实现是错误的。您必须返回 temp （未更改的对象）而不是修改的对象：

Your implementation of the postfix operator is wrong. You must return temp (the unchanged object) instead of the modified object:

Time operator++(int)
{
    Time temp(hours, minutes);
    minutes++;
    if (minutes >= 60)
    {
        hours++;
        minutes = minutes - 60;
    }
    // Return unchanged object here!
    return temp;
}

另请参阅增量和递减运算符重载（C ++） [ ^ ]。

使用后缀实现中的前缀增量：

See also Increment and Decrement Operator Overloading (C++)[^].
That uses the prefix increment within the postfix implementation:

Time operator++(int)
{
    Time temp = *this;
    *++this;
    return temp;
}

由于您要覆盖两个运算符，因此不适用正常的排序规则。在将结果传递到左侧之前，将完全评估每个表达式的右侧。使用调试器逐步执行代码将显示发生的情况。

Since you are overriding both operators the normal ordering rules do not apply. The right-hand side of each expression will be completely evaluated before the result is passed to the left-hand side. Stepping through the code with your debugger will show you what happens.

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