From: Linus Torvalds > Sent: 24 July 2024 21:03 > > On Wed, 24 Jul 2024 at 09:49, Arnd Bergmann <arnd@xxxxxxxxxx> wrote: > > > > I don't understand why this return '0' for unsigned types, > > shouldn't this be > > > > ((is_unsigned_type(typeof(x)) ? 1 : __if_constexpr(x, (x) + 0, -1)) >= 0) > > Yes, that looks more logical. The condition is '>= 0' so it doesn't matter if it is '1' or '0'. > Plus why do that "__if_constexpr(x, (x) + 0, -1)) >= 0)" when it would > appear to be more logical to move the comparison inside, ie > > __if_constexpr(x, (x) >= 0, 0) That gives a 'comparison of unsigned type against 0 is always true' warning. (The compiler generates that for code in the unused branches of both __builtin_choose_expr() and _Generic().) Moving the comparison to the outer level stops all such compiler warnings. > but I also don't see why that "+ 0" existed in the original. So > there's presumably something I'm missing. IIRC it was there to convert a 'bool' to 'int'. Somewhere the is a max(bool,bool) that could just be |. If may not be needed now the expansion is '(cond ? 0 : bool) >= 0' since the 'bool' picks up an (int) cast from the result of ?:. David > I do get the feeling that the problem came from us being much too > clever with out min/max macros, and now this series is doubling down > instead of saying "it wasn't really worth it". > > Linus - Registered Address Lakeside, Bramley Road, Mount Farm, Milton Keynes, MK1 1PT, UK Registration No: 1397386 (Wales)
