David Laight <david.lai...@aculab.com> 於 2022年6月1日 週三 下午4:25寫道：
>
> From: Yu-Jen Chang
> > Sent: 01 June 2022 06:59
> >
> > David Laight <david.lai...@aculab.com> 於 2022年5月30日 週一 下午4:10寫道：
> > >
> > > From: Yu-Jen Chang
> > > > Sent: 28 May 2022 09:13
> > > >
> > > > The original assembly version of memchr() is implemented with
> > > > the byte-wise comparing technique, which does not fully
> > > > use 64-bits registers in x86_64 CPU. We use word-wide
> > > > comparing so that 8 characters can be compared at the same time
> > > > on x86_64 CPU. First we align the input and then use word-wise
> > > > comparing to find the first 64-bit word that contain the target.
> > > > Secondly, we compare every byte in the word and get the output.
> > > >
> > > > We create two files to measure the performance. The first file
> > > > contains on average 10 characters ahead the target character.
> > > > The second file contains at least 1000 characters ahead the
> > > > target character. Our implementation of “memchr()” is slightly
> > > > better in the first test and nearly 4x faster than the orginal
> > > > implementation in the second test.
> > > >
> > > > Signed-off-by: Yu-Jen Chang <arthurchan...@gmail.com>
> > > > Signed-off-by: Ching-Chun (Jim) Huang <js...@ccns.ncku.edu.tw>
> > > > ---
> > > > arch/x86/include/asm/string_64.h | 3 ++
> > > > arch/x86/lib/Makefile | 1 +
> > > > arch/x86/lib/string_64.c | 78 ++++++++++++++++++++++++++++++++
> > > > 3 files changed, 82 insertions(+)
> > > > create mode 100644 arch/x86/lib/string_64.c
> > > >
> > > ...
> > > > diff --git a/arch/x86/lib/string_64.c b/arch/x86/lib/string_64.c
> > > > new file mode 100644
> > > > index 000000000..4e067d5be
> > > > --- /dev/null
> > > > +++ b/arch/x86/lib/string_64.c
> > > > @@ -0,0 +1,78 @@
> > > > +// SPDX-License-Identifier: GPL-2.0
> > > > +#include <linux/string.h>
> > > > +#include <linux/export.h>
> > > > +#include <linux/align.h>
> > > > +
> > > > +/* How many bytes are loaded each iteration of the word copy loop */
> > > > +#define LBLOCKSIZE (sizeof(long))
> > > > +
> > > > +#ifdef __HAVE_ARCH_MEMCHR
> > > > +
> > > > +void *memchr(const void *cs, int c, size_t length)
> > > > +{
> > > > + const unsigned char *src = (const unsigned char *)cs, d = c;
> > >
> > > You don't need the cast.
> > >
> > > > +
> > > > + while (!IS_ALIGNED((long)src, sizeof(long))) {
> > > > + if (!length--)
> > > > + return NULL;
> > > > + if (*src == d)
> > > > + return (void *)src;
> > > > + src++;
> > > > + }
> > >
> > > There is no point aligning the address.
> > > On tests I've done misaligned reads don't even take an extra
> > > clock - even if you get the cpu doing two reads/clock.
> > > Even if they did the code isn't memory limited.
> > >
> > > > + if (length >= LBLOCKSIZE) {
> > > > + unsigned long mask = d << 8 | d;
> > > > + unsigned int i = 32;
> > > > + long xor, data;
> > > > + const long consta = 0xFEFEFEFEFEFEFEFF,
> > > > + constb = 0x8080808080808080;
> > > > +
> > > > + /*
> > > > + * Create a 8-bytes mask for word-wise comparing.
> > > > + * For example, a mask for 'a' is 0x6161616161616161.
> > > > + */
> > > > +
> > > > + mask |= mask << 16;
> > > > + for (i = 32; i < LBLOCKSIZE * 8; i <<= 1)
> > > > + mask |= mask << i;
> > >
> > > Given that consta/b only support 64 bit why the loop.
> > > Just do mask |= mask << 32.
> > > I'd also put all 3 calculations together - not hide one
> > > in the initialiser.
> > >
> > > > + /*
> > > > + * We perform word-wise comparing with following
> > > > operation:
> > > > + * 1. Perform xor on the long word @src and @mask
> > > > + * and put into @xor.
> > > > + * 2. Add @xor with @consta.
> > > > + * 3. ~@xor & @constb.
> > > > + * 4. Perform & with the result of step 2 and 3.
> > > > + *
> > > > + * Step 1 creates a byte which is 0 in the long word if
> > > > + * there is at least one target byte in it.
> > > > + *
> > > > + * Step 2 to Step 4 find if there is a byte with 0 in
> > > > + * the long word.
> > > > + */
> > > > + asm volatile("1:\n\t"
> > > > + "movq (%0),%1\n\t"
> > > > + "xorq %6,%1\n\t"
> > > > + "lea (%1,%4), %2\n\t"
> > > > + "notq %1\n\t"
> > > > + "andq %5,%1\n\t"
> > > > + "testq %1,%2\n\t"
> > > > + "jne 2f\n\t"
> > > > + "add $8,%0\n\t"
> > > > + "sub $8,%3\n\t"
> > > > + "cmp $7,%3\n\t"
> > > > + "ja 1b\n\t"
> > > > + "2:\n\t"
> > > > + : "=D"(src), "=r"(xor), "=r"(data),
> > > > "=r"(length)
> > >
> > > Why constrain src to %rdi?
> >
> > At first I try to use some instructions related to %rdi, but I realize
> > that I won't use these instructions. It is unnecessary to constrain
> > src to %rdi.
> >
> > >
> > > > + : "r"(consta), "r"(constb), "r"(mask),
> > > > "0"(src),
> > > > + "1"(xor), "2"(data), "3"(length)
> > >
> > > Use "+r" in the outputs instead of respecifying the args.
> > > I'd also suggest using named arguments - much easier to read.
> > >
> > > > + : "memory", "cc");
> > >
> > > Doesn't the compiler generate much the same code?
> > > You should also be able to code without needing add, sub and cmp
> > > at the end of the loop.
> > > If you use negative offsets from the end of the buffer
> > > the loop can be a single add and jnz.
> > >
> > > David
> > >
> > > > + }
> > > > +
> > > > + while (length--) {
> > > > + if (*src == d)
> > > > + return (void *)src;
> > > > + src++;
> > > > + }
> > > > + return NULL;
> > > > +}
> > > > +EXPORT_SYMBOL(memchr);
> > > > +#endif
> > > > --
> > > > 2.25.1
> > >
> > > -
> > > Registered Address Lakeside, Bramley Road, Mount Farm, Milton Keynes, MK1
> > > 1PT, UK
> > > Registration No: 1397386 (Wales)
> >
> > I remove the aligning address part. On my tests the performance are similar.
> > Here I rewrite the assembly using named arguments and I reduce one
> > instruction
> > in the loop by adding two parameters, which are 'end' and 'dst'.
> > 'end' stores the
> > address of the end of the string. 'dst' stores the address of the end
> > of word-wise
> > comparison. As a result, when 'src' is equal to 'dst', the number of
> > remaining
> > characters is less than 8. The following while loop will find if the
> > target character is
> > in these remaining characters.
> >
> > On my test the performance is similar with the my original implementation.
> > Only
> > a little bit fast when going through a very long string, which contains
> > 128*1024
> > characters and the target character is near the end of the string.
> >
> > I also explain how consta and constb work clearly in the comments. Hope
> > that it
> > helps understanding.
> >
> > The following code is what I change.
> >
> > void *memchr(const void *cs, int c, size_t length)
> > {
> > const unsigned char *src = (const unsigned char *)cs;
> > const unsigned char *end = src + length;
> >
> > if (length >= LBLOCKSIZE) {
> > unsigned long mask = c << 8 | c;
>
> That is wrong if 'c' is outside 0..255.
> I suspect it is best to at least allow -128..-1.
>
> > long xor, data;
> > const long consta = 0xFEFEFEFEFEFEFEFF,
> > constb = 0x8080808080808080;
> > const unsigned char *dst = (const unsigned char *)src +
> > (length &
> > 0xFFFFFFFFFFFFFFF8);
> >
> > /*
> > * Create a 8-bytes mask for word-wise comparing.
> > * For example, a mask for 'a' is 0x6161616161616161.
> > */
> >
> > mask |= mask << 16;
> > mask |= mask << 32;
> > /*
> > * We perform word-wise comparing with following operation:
> > * 1. Perform xor on the long word @src and @mask
> > * and put into @xor.
> > * 2. Add @xor with @consta.
> > * 3. ~@xor & @constb.
> > * 4. Perform & with the result of step 2 and 3.
> > *
> > * If there is a zero byte in @xor, step 2 turns it into
> > * 0xFF. Then step 3 and 4 turn it into 0x80.
> > *
> > * If there is a none-zero byte in @xor, let k
> > * (0 <= k <= 7) be the lowest 1 in this byte. The lowest
> > * k bits are 0. After step 2, the byte ends in a single
> > * bit of value 0. Step 3 and 4 turns this byte into k
> > * bits of 1, which is 2^k - 1, at first. Then & @constb
> > * makes it into 0.
> > *
> > * Step 2 to Step 4 find if there is a byte with 0 in
> > * the long word.
> > */
> > asm volatile("1:\n\t"
> > "movq (%[src]),%[xor]\n\t"
> > "xorq %[mask],%[xor]\n\t"
> > "lea (%[xor],%[const_a]), %[tmp]\n\t"
> > "notq %[xor]\n\t"
> > "andq %[const_b],%[xor]\n\t"
> > "testq %[xor],%[tmp]\n\t"
> > "jnz 2f\n\t"
> > "add $8,%[src]\n\t"
> > "cmp %[src], %[dst]\n\t"
> > "ja 1b\n\t"
> > "2:\n\t"
> > :
> > [src] "+r"(src), [xor] "+r"(xor), [tmp]
> > "+r"(data)
> > : [const_a] "r"(consta), [const_b] "r"(constb),
> > [mask] "r"(mask), [dst] "r"(dst)
> > : "memory", "cc");
> > }
> >
> > while (src <= end) {
> > if (*src == d)
>
> I think you mean 'c'.
>
> > return (void *)src;
> > src++;
> > }
> > return NULL;
> > }
> >
> > Thanks,
> > Yu-Jen Chang
>
> Gcc compiles this C to the same loop and is easier to read.
> Valid on all LE 64bit systems.
>
> void *memchr(const void *p, int c, unsigned long length)
> {
> unsigned long mask, val;
> const void *end = p + length;
>
> c &= 0xff;
> if (p <= end - 8) {
> mask = c | c << 8;
> mask |= mask << 16;
> mask |= mask << 32;
>
> for (; p <= end - 8; p += 8) {
> val = *(unsigned long *)p ^ mask;
> if ((val + 0xfefefefefefefeffu) & (~val & 0x8080808080808080u))
> break;
> }
> }
>
> for (; p < end; p++)
> if (*(unsigned char *)p == c)
> return p;
Here I think we should return (void*) p. So that there is no
compilation warning.
>
> return NULL;
> }
>
> See https://godbolt.org/z/6rqTqfEsx
>
> David
>
> -
> Registered Address Lakeside, Bramley Road, Mount Farm, Milton Keynes, MK1
> 1PT, UK
> Registration No: 1397386 (Wales)
Your modification is easier to understand. I add the comments to
explain how it work.
I will use it in the second version of the patch. Thanks for your advice.
Yu-Jen Chang
void *memchr(const void *p, int c, unsigned long length)
{
unsigned long mask, val;
const void *end = p + length;
c &= 0xff;
if (p <= end - 8) {
/*
* Create a 8-bytes mask for word-wise comparing.
* For example, a mask for 'a' is 0x6161616161616161.
*/
mask = c | c << 8;
mask |= mask << 16;
mask |= mask << 32;
/*
* We perform word-wise comparing with following operation:
* 1. Perform xor on the long word @p and @mask
* and put into @val.
* 2. Add @val with 0xfefefefefefefeff.
* 3. ~@val & 0x8080808080808080
* 4. Perform & with the result of step 2 and 3.
*
* If there is a zero byte in @val, step 2 turns it into
* 0xFF. Then step 3 and 4 turn it into 0x80.
*
* If there is a none-zero byte in @val, let k
* (0 <= k <= 7) be the lowest 1 in this byte. The lowest
* k bits are 0. After step 2, the byte ends in a single
* bit of value 0. Step 3 and 4 turns this byte into k
* bits of 1, which is 2^k - 1, at first. Then &
* 0x8080808080808080 makes it into 0
*/
for (; p <= end - 8; p += 8) {
val = *(unsigned long *)p ^ mask;
if ((val + 0xfefefefefefefeffu) & (~val & 0x8080808080808080u))
break;
}
}
for (; p < end; p++)
if (*(unsigned char *)p == c)
return (void *)p;
return NULL;
}
_______________________________________________
linux-um mailing list
linux-um@lists.infradead.org
http://lists.infradead.org/mailman/listinfo/linux-um
