https://gcc.gnu.org/bugzilla/show_bug.cgi?id=95242
--- Comment #4 from Jonathan Wakely <redi at gcc dot gnu.org> --- It's consteval, the throw is there to make it not a constant expression and give an error if anything except 0 is used. i.e. it can never throw, it either compiles or it doesn't. But I've remembered the problem with this technique, it allows non-literal zeros: constexpr int zero = 0; auto c = (1 <=> 1) == zero; // should be ill-formed. This compiles OK, and isn't *obviously* wrong when reviewing the code. The advantage of the current definition using a pointer is that nobody can declare a constant of type __unspec without obviously using a reserved name and venturing into undefined territory.
