What is meant by non-member operator overloading

Question

I was recently going over an article on operator overloading in which it mentioned non-member operator overloading. I would appreciate it if someone could explain what is meant by non-member operator overloading with an example. I know what member operator overloading is (A method in a class which overloads an operator type (binary..etc) . I came across this post on SO which makes me believe that the purpose of non-member operator overloading is to handle operator overloading in which the first parameter is a not a class and is simply a native type. Any links or examples that explain what non-member operator overloading is would definitely be appreciated.

The link you gave shows an example for non member operator overloading, used as you said to perform add on a primitive — omer schleifer, Mar 17 '13 at 21:57
[This](http://stackoverflow.com/q/4421706/1410711) might be helpful... — Recker, Mar 17 '13 at 21:57

sehe · Accepted Answer · 2013-03-17T22:08:36.023

It means you can overload out-of-class:

struct X { int data; };

bool operator<(X const& a, X const& b) 
{
       return a.data < b.data;
}

This is useful for assymetrical overloading, where the left operand doesn't need to be your own type:

bool operator<(int a, X const& b) 
{
       return a < b.data;
}

A common idiom here is to combine it with in-class definition and friend declaration:

struct X 
{ 
    int data; 
    friend bool operator<(X const& a, X const& b) { return a.data<b.data; }
    friend bool operator<(int a, X const& b) { return a<b.data; }
};

Here, operator< is still technically non-member.

Another useful side-effect of this, as pointed out by DrewDormann below, is that the (X const&, X const&) will apply to any operands that are implicitly convertible to X const&, not just expressions of that exact type.

It's also useful for *symetrical* type conversion, allowing implicit type conversions for all involved operands. — Drew Dormann, Mar 17 '13 at 21:59

score 0 · Answer 2 · answered Mar 17 '13 at 22:05

the most common way is to overload operator<< that will be called on std::cout:

namespace X { 
      class MyClass { 
       ...
       };  
}
std::ostream& operator<< (std::ostream&, const X::MyClass&);

this is called on std::ostream member so you don't define it inside your class. however sometimes the functionality cannot be achieved via the public interfaces (because your operator needs access to data representation).

What is meant by non-member operator overloading

2 Answers2