问题描述
标准是否保证std::less<MyEnumType>将对MyEnumType进行排序，就像将MyEnumType的值强制转换为适当大小的整数类型一样?
Does the standard guarantee that std::less<MyEnumType> will order MyEnumType as if a value of MyEnumType was cast to an appropriately sized integer type?
enum MyEnumType { E1 = 0, E2 = 6, E3 = 3 };
推荐答案
是的，std::less::operator()被定义为(§20.8.5/5):
Yes, std::less::operator() is defined as (§20.8.5/5):
对于在枚举类型上使用关系运算符，声明如下(第5.9/2节):
For using relational operators on enumeration types, the following is stated (§5.9/2):
对于无范围的枚举类型，通常的算术转换被定义为进行积分提升.无作用域枚举类型的整体提升定义为(§5/9):
For unscoped enumeration types, the usual arithmetic conversions are defined as doing integral promotion. Integral promotion for unscoped enumeration types is defined as (§5/9):
如果有需要，将使用扩展的整数类型.
An extended integer type will be used if available and required.
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