std ::更少的枚举 [英] std::less on enums

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):

operator()返回x < y

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):

通常的算术转换是对算术或枚举类型的操作数执行的.

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

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):

其基础类型不固定(7.2)的无范围枚举类型的prvalue可以转换为可以表示该枚举的所有值的以下类型的第一个prvalue，即范围b中的值 _min 到b _max ，如7.2中所述):int，unsigned int，long int，unsigned long int，long long int或unsigned long long int. /p>

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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