c++ - 内置运算符候选人

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."

我不确定从标准中引用这句话的意图。是否可以定义与内置运算符具有相同名称和类型的用户定义候选函数？

例如以下显然是错误的。

int operator+(int)

那么这句话是什么意思？

Answer 1

只需在 13.6 中选择一个即可。喜欢

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);

所以

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