我正在尝试运算符重载,发现了一些我无法解释的东西：
WeekDays.h
using namespace std;
enum DAYS
{
MON,TUE,WED,THU,FRY,SAT,SUN
};
DAYS operator+(DAYS&a,DAYS &b)
{
printf("Binary+ called\n");
return (DAYS)(((unsigned int)a+(unsigned int)b)%7);
}
//Increment 3
DAYS operator+(DAYS&a)
{
printf("Unary+ called\n");
return (DAYS)(((unsigned int)a+3)%7);
}
ostream& operator<<(ostream&o,DAYS &a)
{
switch(a){
case MON: o<<"MON"; break;
case TUE: o<<"TUE"; break;
case WED: o<<"WED"; break;
case THU: o<<"THU"; break;
case FRY: o<<"FRY"; break;
case SAT: o<<"SAT"; break;
case SUN: o<<"SUN"; break;
}
return o;
};
Main.cpp的
#include <iostream>
#include "WeekDays.h"
using namespace std;
void main()
{
DAYS a=MON; //=0
DAYS b=TUE; //=1
cout<< +a <<endl;
cout<< +b <<endl;
cout<< +(a,b) <<endl;
cout<< (a+b) <<endl;
cin.get();
}
输出是
Unary+ called 3 Unary+ called 4 Unary+ called 4 Binary+ called 1
为什么(a,b)被评估为一元运算符b？我没有解释这一点.
链接到相关主题Operator overloading.
我正在使用VisualStudio 2012.
解决方法
使用(a,b),您恰好调用奇数
“comma” operator,首先计算a,然后计算b,最后返回b.
您可以通过将其拼写为运算符(a,b)来调用运算符. (这里的逗号是参数的分隔符,而不是逗号运算符).
