Question

C++03 $13.6/1- “[…]If there is a
user-written candidate with the same
name and parameter types as a built-in
candidate operator function, the
built-in operator function is hidden
and is not included in the set of
candidate functions.”

I am not sure about the intent of this quote from the Standard. Is it possible to define a user defined candidate function that has the same name and type as a built-in operator?

e.g. the below which is clearly wrong.

int operator+(int)

So what does this quote mean?

Answer 1

Just pick one of those in 13.6. Like

For every pointer or enumeration type T, there exist candidate operator functions of the form

bool operator<(T, T);
bool operator>(T, T);
bool operator<=(T, T);
bool operator>=(T, T);
bool operator==(T, T);
bool operator!=(T, T);

So

enum Kind { Evil, Good };
bool operator<(Kind a, Kind b) { ... }