c++ - std::less 枚举

Question

标准是否保证std::less<MyEnumType>会订MyEnumType好像值为 MyEnumType被转换为适当大小的整数类型？

enum MyEnumType { E1 = 0, E2 = 6, E3 = 3 };

Answer 1

是的，std::less::operator()定义为 (§20.8.5/5):

operator() returns x < y

对于在枚举类型上使用关系运算符，声明如下 (§5.9/2):

The usual arithmetic conversions are performed on operands of arithmetic or enumeration type.

对于无作用域的枚举类型，通常的算术转换 被定义为进行整数提升。无范围枚举类型的整体提升定义为 (§5/9):

A prvalue of an unscoped enumeration type whose underlying type is not fixed (7.2) can be converted to a prvalue of the first of the following types that can represent all the values of the enumeration (i.e., the values in the range b_min to b_max as described in 7.2): int, unsigned int, long int, unsigned long int, long long int, or unsigned long long int.

如果可用且需要，将使用扩展整数类型。