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Message-ID: <85p0oop-5pq-p6o-7560-297sn1np3os@syhkavp.arg>
Date: Mon, 30 Nov 2020 12:52:25 -0500 (EST)
From: Nicolas Pitre <nico@...xnic.net>
To: Ard Biesheuvel <ardb@...nel.org>
cc: Antony Yu <swpenim@...il.com>,
Nick Desaulniers <ndesaulniers@...gle.com>,
Russell King <linux@...linux.org.uk>,
Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
clang-built-linux <clang-built-linux@...glegroups.com>,
Nathan Chancellor <natechancellor@...il.com>,
Linux ARM <linux-arm-kernel@...ts.infradead.org>
Subject: Re: [RESEND,PATCH] ARM: fix __div64_32() error when compiling with
clang
On Mon, 30 Nov 2020, Ard Biesheuvel wrote:
> On Mon, 30 Nov 2020 at 16:51, Nicolas Pitre <nico@...xnic.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.
Nicolas
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