std::less on enums
c++, enums, language-lawyer
Solution
Yes, `std::less::operator()` is defined as (§20.8.5/5):
`operator()` returns `x < y`
For using relational operators on enumeration types, the following is stated (§5.9/2):
The usual arithmetic conversions are performed on operands of arithmetic or enumeration type.
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):
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 bmin to bmax as described in 7.2): `int`, `unsigned int`, `long int`, `unsigned long int`, `long long int`, or `unsigned long long int`.
An extended integer type will be used if available and required.
Problem
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 }; ```