C++操作符重载总结

/**
    C++操作符重载时,首先找成员函数是否已经有相关实现,如果有相关的实现,则优先调用成员函数的操作符重载,如果没有的话,则调用全局的操作符重载.
    总结:
    1) 不可以重载的操作符 (1) :: (2) .*  (3) ?:
    2) 重载操作符必须有一个类类型或枚举类型,意思是不能重载内置的基本数据类型的操作符
    3) 优先级和结合性是固定的
    4) 不保证操作符的调用顺序
    5) 类成员函数和非类成员函数: 比较操作符适用于非类成员,赋值操作适用于类成员.如果对于同一个操作,既定义了member function,也定义非member function,则member operator overwrite function优先级高
    6) 对于none member function的操作符重载,往往需要声明为friend,以便访问private data.
    7) 可以使用操作符重载显示调用
        如 item1 + item2; //优先调用member function,若无实现,则调用none member function
        等价于 operator + ( iterm1,item2)
    author : openxmpp@163.com
    date 2013-2-23
*/
    
#include <iostream>
using namespace std;
#include <string>
using std::string;
class MyClass
{
    friend bool operator < ( const MyClass &value1, const MyClass &value2 );
    friend bool operator == ( const MyClass &value1, const MyClass &value2 );
public:
    explicit MyClass( int value )
    {
        mNumber = value;
    }
    bool operator < ( const MyClass & value )
    {
        cout << "member function operator < called " << mNumber << "  " << value.mNumber<<endl;
        return this->mNumber < value.mNumber;
    }
   bool operator == ( const MyClass & value )
   {
       cout << "member function operator == called " << mNumber << "\t" << value.mNumber<<endl;
       return this->mNumber ==  value.mNumber;
   }
   MyClass & operator + ( const MyClass & rvalue )
   {
       cout << "member function operator + called "<< mNumber << "\t" << rvalue.mNumber<<endl;
       mNumber += rvalue.mNumber;
       return *this;
   }
private:
    int mNumber;
};

bool operator < ( const MyClass &lValue, const MyClass &rValue )
{
    cout << "operator global operator < called"<<endl;
    return lValue.mNumber == rValue.mNumber;
}

bool operator == ( const MyClass &lValue,const MyClass &rValue )
{
    cout << "operator global operator == called"<<endl;
    return lValue.mNumber == rValue.mNumber;
}

//这将导致编译错误,不能重载内置的基本数据类型的操作符
#if 0
bool  operator == ( int lValue,int rValue)
{
    return lValue < rValue;
}
#endif

int main(int argc, char *argv[])
{
    MyClass c(1024);
    MyClass d(2048);
    c + d; //调用MyClass的 operator + ,这里期望是返回的一个临时变量,但结果是在原地修改c的值了,违背设计初衷,因此,对于累加的操作符,建议定义为全局的
    operator == ( c,d ); //bool operator == ( const MyClass &lValue,const MyClass &rValue )
    d == c + c; //优先级和结合性是固定的,先是"+",再是"=="
    return 0;
}