On 2025-04-25 at 00:38, Vladimir Grigoriev via Std-Discussion wrote:
> In the C++20 Standard there is written (9.7.1 Enumeration declarations):
> 7 For an enumeration whose underlying type is not fixed, the underlying
> type is an integral type that can represent all the enumerator values
> defined in the enumeration. If no integral type can represent all the
> enumerator values, the enumeration is ill-formed. It is implementation-
> defined which integral type is used as the underlying type *except that
> the underlying type shall not be larger than int unless the value of an
> enumerator cannot fit in an int or unsigned int*. If the enumerator-list
> is empty, the underlying type is as if the enumeration had a single
> enumerator with value 0.
> However if to run this simple program
>
> |#include <iostream> #include <type_traits> int main() { int vals[]
> { 2,3,5,7,11 }; enum { N = sizeof vals / sizeof vals[0] }; std::cout <<
> std::boolalpha; std::cout <<
> "std::is_same_v<std::underlying_type_t<decltype( N )>, size_t> = " <<
> std::is_same_v<std::underlying_type_t<decltype( N )>, size_t> << '\n'; }|
>
> |then the MS VS2022 outputs *true *though size_t is an alias for the
> type unsigned long long and the constant value used as an initializer
> of the enumerator *N *at least can fit in an object of the type
> unsigned int.|

For me it outputs false, and the IDE popup says that the type is int.