Operator overloading outside class [duplicate]

Answer 1

There is an excellent discussion of this issue in Meyer's Effective C++: Item 24 is "Declare non-member functions when type conversions should apply to all parameters" and Item 46 is "Define non-member functions inside templates when type conversions are desired".

Answer 2

The basic question is "Do you want conversions to be performed on the left-hand side parameter of an operator?". If yes, use a free function. If no, use a class member.

For example, for operator+() for strings, we want conversions to be performed so we can say things like:

string a = "bar";
string b = "foo" + a;

where a conversion is performed to turn the char * "foo" into an std::string. So, we make operator+() for strings into a free function.

Answer 3

First: the two different ways are really "overload as a member" and "overload as a non-member", and the latter has two different ways to write it (as-friend-inside class definition and outside class definition). Calling them "inside class" and "outside class" is going to confuse you.

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Overloads for +=, +, -=, -, etc. have a special pattern:

struct Vector2 {
  float x, y;
  Vector2& operator+=(Vector2 const& other) {
    x += other.x;
    y += other.y;
    return *this;
  }
  Vector2& operator-=(Vector2 const& other) {
    x -= other.x;
    y -= other.y;
    return *this;
  }
};
Vector2 operator+(Vector2 a, Vector2 const& b) {
  // note 'a' is passed by value and thus copied
  a += b;
  return a;
}
Vector2 operator-(Vector2 a, Vector2 const& b) { return a -= b; } // compact

This pattern allows the conversions mentioned in the other answers for the LHS argument while simplifying the implementation considerably. (Either member or non-member allows conversions for the RHS when it's passed either as a const& or by value, as it should be.) Of course, this only applies when you do actually want to overload both += and +, -= and -, etc., but that is still common.

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Additionally, you sometimes want to declare your non-member op+, etc. as friends within the class definition using the Barton-Nackman trick, because due to quirks of templates and overloading, it may not be found otherwise.