重载+ =运算符作为朋友函数

Question

I read that implementing operators as friend function will be better. How to overload += operator as a friend function when I already have + operator function:

friend Dollar operator+(const Dollar &p1, const Dollar &p2);
friend Dollar &operator+=(const Dollar &p1, const Dollar &p2);

This is wrong since I need to return a reference to a variable.

Dollar &operator+=(const Dollar &p1, const Dollar &p2)
{
    return p1+p2;
}

Answer 1

Overloading operators as friend functions is better in order to allow conversions to apply to both the left and right side of the expression. For example, string 's operator+ is a friend so that I can write "Hello " + string("World") rather than only being able to write string("Hello ") + "World" .

However, this reasoning doesn't apply to mutators such as operator+= . You have to take a non-const left argument, which precludes being able to use this operator on a temporary. For this reason, it's recommended to implement non-mutating operators as friend (or otherwise free-) functions and mutators as member functions. (In fact, operator= can only be overloaded as a member function.)

Answer 2

You also need to modify the lvalue.

Dollar &operator+=(Dollar &p1, const Dollar &p2)
{
    p1=p1+p2;
    return p1;
}

Answer 3

通常的方法是提供+=作为修改*this的成员函数，并将+实现为使用+=的自由函数。