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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T17:26:15+00:00 2026-06-15T17:26:15+00:00

Does the standard guarantee that std::less<MyEnumType> will order MyEnumType as if a value of

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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 };
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  1. Editorial Team
    Editorial Team
    2026-06-15T17:26:15+00:00Added an answer on June 15, 2026 at 5:26 pm

    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.

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