According to the C++23 standard, what is the output of this program?
#include <iostream>
int main() {
constexpr unsigned int id = 100;
unsigned char array[] = { id % 3, id % 5 };
std::cout << static_cast<unsigned int>(array[0])
<< static_cast<unsigned int>(array[1]) ;
}Would the result be different of we removed constexpr from the declaration of id?
10
array is an array of unsigned char, but is initialized with unsigned ints. The unsigned ints have to be converted to unsigned chars, and since narrowing is not allowed in aggregate-initialization from an initializer list, we might expect a compilation error.
When an aggregate is initialized by an initializer list as specified in §[dcl.init.list], the elements of the initializer list are taken as initializers for the elements of the aggregate.
And §[dcl.init.aggr]¶4:
For each explicitly initialized element (...) [if] a narrowing conversion §[dcl.init.list] is required to convert the expression, the program is ill-formed.
§[dcl.init.list]¶7 tells us what a narrowing conversion is:
A narrowing conversion is an implicit conversion (...) from an integer type (...) to an integer type that cannot represent all the values of the original type (...).
An unsigned char cannot represent all the values of an unsigned int, so we would normally expect a diagnostic. However, the above clause continues with an exception for the case where the source is a constant expression whose value after promotion fits in the target type:
(...) except where (...) the source is a constant expression whose value after integral promotions will fit into the target type (...).
So since id is constexpr, and both id % 3 and id % 5 fit in an unsigned char, the program is well-formed.