On Fri, 2008-04-11 at 19:09 +0100, Russell King wrote:
> On Thu, Apr 10, 2008 at 05:06:34PM -0700, Harvey Harrison wrote:
> > +#define __get_unaligned_cpu(ptr) ({ \
> > + const void *__gu_p = (ptr); \
> > + typeof(*(ptr)) __val; \
> > + switch (sizeof(*(ptr))) { \
> > + case 1: \
> > + __val = *(const u8 *)__gu_p; \
> > + break; \
> > + case 2: \
> > + __val = __get_unaligned_cpu16(__gu_p); \
> > + break; \
> > + case 4: \
> > + __val = __get_unaligned_cpu32(__gu_p); \
> > + break; \
> > + case 8: \
> > + __val = __get_unaligned_cpu64(__gu_p); \
> > + break; \
> > + default: \
> > + BUILD_BUG_ON(1); \
> > + break; \
> > + }; \
> > + __val; })
>
> This won't work - on ARM we used to use this style, but it fails in
> some corner case, so we ended up switching to using GCC's
> __builtin_choose_expr() instead.
>
> Such a corner case:
>
> static unsigned long foo(const unsigned long *ptr)
> {
> return __get_unaligned_cpu(ptr);
> }
>
> This results in '__val' being declared as const, and therefore the
> compiler errors out in the switch statement since the code tries to
> assign to a const '__val'.
OK, will respin the set, I think I have a better way of doing it anyway.
Not that it matters, but wouldn't that be a gcc bug, for it to preserve
const on __val, wouldn't that have to be:
const unsigned long * const ptr
Just curious.
Thanks.
Harvey
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