From: James Bottomley > Sent: 25 October 2018 16:33 > > On Thu, 2018-10-25 at 16:13 +0100, Colin King wrote: > > From: Colin Ian King <colin.king@xxxxxxxxxxxxx> > > > > In the expression "ahc_inb(ahc, port+3) << 24", the initial value is > > a u8, but is promoted to a signed int, then sign-extended to > > uint64_t. > > Why is this, that's highly non intuitive? The compiler is supposed to > promote to the biggest type, which is uint64_t and then do the > calculation Do not doubt the wisdom on the ANSI C committee that decided to do 'value preserving' integer promotions instead of the 'sign preserving' ones of K&R C. So 'unsigned char' is promoted to 'int' almost everywhere it is used (unless they are both the same size - which is allowed). This means that ahc_inb() << 24 is actually undefined (signed integer overflow can do anything it likes). By far the best fix is to change the return type of ahc_inb() to be 'unsigned int'. On systems without byte sized registers (about everything except x86) this will almost certainly generate better code. David - Registered Address Lakeside, Bramley Road, Mount Farm, Milton Keynes, MK1 1PT, UK Registration No: 1397386 (Wales)
