Overloading operator= as Non-Member [duplicate]

This question already has answers here:开发者_Go百科 Why cannot a non-member function be used for overloading the assignment operator? (9 answers) Closed 8 years ago.

According to replies to this thread, operator= cannot be overloaded as a non-member function. So, for example, the following makes the compiler very angry:

class MyClass
{
    // ...
};

MyClass& operator=(MyClass& Left, MyClass& Right)
{
    // ...
}

Why is this? I have a container class with getters and setters, so a member function is unnecessary and it would break encapsulation. One of the answers to the aforementioned thread said that it's to make sure the "L-value is received as its first operand," but I don't fully understand what that means. Could someone please clarify?

Additionally, are operator=, operator(), operator[] and operator-> "oddball" cases...? Or should I implement all overloaded operators as member functions...? (I know it's perfectly legal to do otherwise, but I'm looking for the better practice.)

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If your class doesn't have an assignment operator (as a member), the compiler generates one by default, just like it generates a copy constructor if you don't provide one.

Therefore it will be "angry" if you try to define a non-member assignment operator later. There will then be two!

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Why is this?

Unless you declare one, the compiler declares an operator= in your class with signature operator= (C&, C&) or operator= (C&, const C&).

If you were allowed to overload assignment, most uses would be ambiguous.

Then you would likely lobby for additional rules to either:

• pretend there is no compiler declared operator= if a user declared one is visible, as if the non-member operator= hides the member operator=
• make your user declared assignment a better match during overloading.

Both choices would complicate rules that are already extremely complicated, adding a special case just for operator=.

Few people want to go there.

And also you haven't exposed a legitimate use for this feature.

Or, any reasonable use at all.

The C++ rules are only made even more complicated when some reasonable use-case can be showed.