C运算符重载称为函数

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).