https://gcc.gnu.org/bugzilla/show_bug.cgi?id=103096
--- Comment #9 from Jonathan Wakely <redi at gcc dot gnu.org> ---
(In reply to Jonathan Wakely from comment #5)
> (In reply to Jonathan Wakely from comment #4)
> > Maybe the compiler could notice that there is only one qHash function
> > template, and so realize that the instantiation recursion of qHash<X>
> > instantiating qHash<Y> instantiating qHash<Z> etc. will never terminate.
>
> It would also have to consider whether there are partial or explicit
> specializations of the pair class template, because otherwise one of them
> could define a friend that would be found by ADL and terminate the
> recursion. So maybe too hard.
And if-constexpr means you can't just detect the recursion based on whether
name lookup finds more than one candidate:
template <typename T> int qHash(const T& a)
{
if constexpr (delved too deep)
return balrog;
return qHash(pair<T, T>(a, a));
}
This one will terminate, even though it keeps calling itself.
