Return the underlying
or throw an exception.
std::uint64_t as_uint64() const;
true, returns the underlying
std::uint64_t, otherwise throws an exception.
This function is the const-qualified overload of
as_uint64, which is intended
for direct access to the underlying object, if
it has the type
std::uint64_t. It does not convert the underlying
object to type
std::uint64_t even if a lossless conversion
is possible. If you are not sure which kind your
has, and you only care about getting a
number, consider using