Operator overloading : member function vs. non-member function?

Question

I read that an overloaded operator declared as member function is asymmetric because it can have only one parameter and the other parameter passed automatically is

Answer 1

It's not necessarily a distinction between friend operator overloads and member function operator overloads as it is between global operator overloads and member function operator overloads.

One reason to prefer a global operator overload is if you want to allow expressions where the class type appears on the right hand side of a binary operator. For example:

Foo f = 100;
int x = 10;
cout << x + f;

This only works if there is a global operator overload for

Foo operator + (int x, const Foo& f);

Note that the global operator overload doesn't necessarily need to be a friend function. This is only necessary if it needs access to private members of Foo, but that is not always the case.

Regardless, if Foo only had a member function operator overload, like:

class Foo
{
  ...
  Foo operator + (int x);
  ...
};

...then we would only be able to have expressions where a Foo instance appears on the left of the plus operator.