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.k...@canonical.com>
> >
> > 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)
