[RESEND, PATCH] ARM: fix __div64_32() error when compiling with clang

[Ard Biesheuvel]

On Mon, 30 Nov 2020 at 18:52, Nicolas Pitre <nico at fluxnic.net> wrote:
>
> On Mon, 30 Nov 2020, Ard Biesheuvel wrote:
>
> > On Mon, 30 Nov 2020 at 16:51, Nicolas Pitre <nico at fluxnic.net> wrote:
> >
> > > Here's my version of the fix which should be correct. Warning: this
> > > is completely untested, but should in theory produce the same code on
> > > modern gcc.
> > >
> > > diff --git a/arch/arm/include/asm/div64.h b/arch/arm/include/asm/div64.h
> > > index 898e9c78a7..595e538f5b 100644
> > > --- a/arch/arm/include/asm/div64.h
> > > +++ b/arch/arm/include/asm/div64.h
> > > @@ -21,29 +21,20 @@
> > >   * assembly implementation with completely non standard calling convention
> > >   * for arguments and results (beware).
> > >   */
> > > -
> > > -#ifdef __ARMEB__
> > > -#define __xh "r0"
> > > -#define __xl "r1"
> > > -#else
> > > -#define __xl "r0"
> > > -#define __xh "r1"
> > > -#endif
> > > -
> > >  static inline uint32_t __div64_32(uint64_t *n, uint32_t base)
> > >  {
> > >         register unsigned int __base      asm("r4") = base;
> > >         register unsigned long long __n   asm("r0") = *n;
> > >         register unsigned long long __res asm("r2");
> > > -       register unsigned int __rem       asm(__xh);
> > > -       asm(    __asmeq("%0", __xh)
> > > +       unsigned int __rem;
> > > +       asm(    __asmeq("%0", "r0")
> > >                 __asmeq("%1", "r2")
> > > -               __asmeq("%2", "r0")
> > > -               __asmeq("%3", "r4")
> > > +               __asmeq("%2", "r4")
> > >                 "bl     __do_div64"
> > > -               : "=r" (__rem), "=r" (__res)
> > > -               : "r" (__n), "r" (__base)
> > > +               : "+r" (__n), "=r" (__res)
> > > +               : "r" (__base)
> > >                 : "ip", "lr", "cc");
> > > +       __rem = __n >> 32;
> >
> > This treats {r0, r1} as a {low, high} pair, regardless of endianness,
> > and so it puts the value of r0 into r1. Doesn't that mean the shift
> > should only be done on little endian?
>
> Not quite. r0-r1 = low-high is for little endian. Then "__n >> 32" is
> actually translated into "mov r0, r1" to move it into __rem and returned
> through r0.
>
> On big endial it is r0-r1 = high-low.  Here "__n >> 32" picks r0 and
> moves it to __rem which is returned through r0 so no extra instruction
> needed.
>
> Of course the function is inlined so r0 can be anything, or optimized
> away if__rem is not used.
>

OK, you're right. I got myself confused there, but a quick test with
GCC confirms your explanation:

$ arm-linux-gnueabihf-gcc -mbig-endian -O2 -S -o - \
   -xc - <<<"long f(long long l) { return l >> 32; }"

just produces

bx lr

whereas removing the -mbig-endian gives

mov r0, r1
bx lr


I tested the change and it builds and runs fine (although I am not
sure how much coverage this code gets on an ordinary boot):

Reviewed-by: Ard Biesheuvel <ardb at kernel.org>
Tested-by: Ard Biesheuvel <ardb at kernel.org>