Consider the example in over.match.oper#7, that is:

struct A {
  operator int();
};
A operator+(const A&, const A&);
void m() {
  A a, b;
  a + b;                        // operator+(a, b) chosen over int(a) + int(b)
}

Note the comment in this code, it sounds like the best overload is built-in operator. However, For this example, the candidate functions are a non-member user-defined function for operator + and a built-in operator, according to [over.match.best], convert these operands of type A to what the type build-in + expects, there are user-defined conversion sequences applied to these original operands, conversely, for user-declared operator +, only identity conversion applied to these operands, which is standard conversion sequences. Obviously, the later is more better. why the comment said chosen over int(a) + int(b)? GCC and Clang all agree that A operator+(const A&, const A&) is the best overload, Is it a typo?