Operator overloading outside class
This question already has an answer here:
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.
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.
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.
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.
在Meyer的Effective C ++中有一个很好的关于这个问题的讨论:第24项是“当类型转换应用于所有参数时声明非成员函数”,而第46项是“在需要类型转换时定义模板内部的非成员函数”。
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