[10/17] string: Improve generic memchr
Commit Message
New algorithm have the following key differences:
- Reads first word unaligned and use string-maskoff function to
remove unwanted data. This strategy follow arch-specific
optimization used on aarch64 and powerpc.
- Use string-fz{b,i} and string-opthr functions.
Checked on x86_64-linux-gnu, i686-linux-gnu, powerpc-linux-gnu,
and powerpc64-linux-gnu by removing the arch-specific assembly
implementation and disabling multi-arch (it covers both LE and BE
for 64 and 32 bits).
Co-authored-by: Richard Henderson <rth@twiddle.net>
---
string/memchr.c | 168 +++++-------------
.../powerpc32/power4/multiarch/memchr-ppc32.c | 14 +-
.../powerpc64/multiarch/memchr-ppc64.c | 9 +-
3 files changed, 48 insertions(+), 143 deletions(-)
Comments
On Fri, Sep 2, 2022 at 1:45 PM Adhemerval Zanella via Libc-alpha
<libc-alpha@sourceware.org> wrote:
>
> New algorithm have the following key differences:
>
> - Reads first word unaligned and use string-maskoff function to
> remove unwanted data. This strategy follow arch-specific
> optimization used on aarch64 and powerpc.
>
> - Use string-fz{b,i} and string-opthr functions.
>
> Checked on x86_64-linux-gnu, i686-linux-gnu, powerpc-linux-gnu,
> and powerpc64-linux-gnu by removing the arch-specific assembly
> implementation and disabling multi-arch (it covers both LE and BE
> for 64 and 32 bits).
>
> Co-authored-by: Richard Henderson <rth@twiddle.net>
> ---
> string/memchr.c | 168 +++++-------------
> .../powerpc32/power4/multiarch/memchr-ppc32.c | 14 +-
> .../powerpc64/multiarch/memchr-ppc64.c | 9 +-
> 3 files changed, 48 insertions(+), 143 deletions(-)
>
> diff --git a/string/memchr.c b/string/memchr.c
> index 422bcd0cd6..8fe0ac48ab 100644
> --- a/string/memchr.c
> +++ b/string/memchr.c
> @@ -1,10 +1,6 @@
> -/* Copyright (C) 1991-2022 Free Software Foundation, Inc.
> +/* Scan memory for a character. Generic version
> + Copyright (C) 1991-2022 Free Software Foundation, Inc.
> This file is part of the GNU C Library.
> - Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
> - with help from Dan Sahlin (dan@sics.se) and
> - commentary by Jim Blandy (jimb@ai.mit.edu);
> - adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
> - and implemented by Roland McGrath (roland@ai.mit.edu).
>
> The GNU C Library is free software; you can redistribute it and/or
> modify it under the terms of the GNU Lesser General Public
> @@ -20,143 +16,65 @@
> License along with the GNU C Library; if not, see
> <https://www.gnu.org/licenses/>. */
>
> -#ifndef _LIBC
> -# include <config.h>
> -#endif
> -
> +#include <intprops.h>
> +#include <string-fza.h>
> +#include <string-fzb.h>
> +#include <string-fzi.h>
> +#include <string-maskoff.h>
> +#include <string-opthr.h>
> #include <string.h>
>
> -#include <stddef.h>
> +#undef memchr
>
> -#include <limits.h>
> -
> -#undef __memchr
> -#ifdef _LIBC
> -# undef memchr
> +#ifdef MEMCHR
> +# define __memchr MEMCHR
> #endif
>
> -#ifndef weak_alias
> -# define __memchr memchr
> -#endif
> -
> -#ifndef MEMCHR
> -# define MEMCHR __memchr
> -#endif
> +static inline const char *
> +sadd (uintptr_t x, uintptr_t y)
> +{
> + uintptr_t ret = INT_ADD_OVERFLOW (x, y) ? (uintptr_t)-1 : x + y;
> + return (const char *)ret;
> +}
>
> /* Search no more than N bytes of S for C. */
> void *
> -MEMCHR (void const *s, int c_in, size_t n)
> +__memchr (void const *s, int c_in, size_t n)
> {
> - /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
> - long instead of a 64-bit uintmax_t tends to give better
> - performance. On 64-bit hardware, unsigned long is generally 64
> - bits already. Change this typedef to experiment with
> - performance. */
> - typedef unsigned long int longword;
> + if (__glibc_unlikely (n == 0))
> + return NULL;
>
> - const unsigned char *char_ptr;
> - const longword *longword_ptr;
> - longword repeated_one;
> - longword repeated_c;
> - unsigned char c;
> + uintptr_t s_int = (uintptr_t) s;
>
> - c = (unsigned char) c_in;
> + /* Set up a word, each of whose bytes is C. */
> + op_t repeated_c = repeat_bytes (c_in);
> + op_t before_mask = create_mask (s_int);
>
> - /* Handle the first few bytes by reading one byte at a time.
> - Do this until CHAR_PTR is aligned on a longword boundary. */
> - for (char_ptr = (const unsigned char *) s;
> - n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
> - --n, ++char_ptr)
> - if (*char_ptr == c)
> - return (void *) char_ptr;
> + /* Compute the address of the last byte taking in consideration possible
> + overflow. */
> + const char *lbyte = sadd (s_int, n - 1);
>
> - longword_ptr = (const longword *) char_ptr;
> + /* Compute the address of the word containing the last byte. */
> + const op_t *lword = word_containing (lbyte);
>
> - /* All these elucidatory comments refer to 4-byte longwords,
> - but the theory applies equally well to any size longwords. */
> + /* Read the first word, but munge it so that bytes before the array
> + will not match goal. */
> + const op_t * word_ptr = word_containing (s);
> + op_t word = (*word_ptr | before_mask) ^ (repeated_c & before_mask);
Why do you xor with repeated_c & before_mask here?
Doesn't the has_eq(word, repeated_c) do that?
>
> - /* Compute auxiliary longword values:
> - repeated_one is a value which has a 1 in every byte.
> - repeated_c has c in every byte. */
> - repeated_one = 0x01010101;
> - repeated_c = c | (c << 8);
> - repeated_c |= repeated_c << 16;
> - if (0xffffffffU < (longword) -1)
> + while (has_eq (word, repeated_c) == 0)
> {
> - repeated_one |= repeated_one << 31 << 1;
> - repeated_c |= repeated_c << 31 << 1;
> - if (8 < sizeof (longword))
> - {
> - size_t i;
> -
> - for (i = 64; i < sizeof (longword) * 8; i *= 2)
> - {
> - repeated_one |= repeated_one << i;
> - repeated_c |= repeated_c << i;
> - }
> - }
> + if (word_ptr == lword)
> + return NULL;
> + word = *++word_ptr;
> }
>
> - /* Instead of the traditional loop which tests each byte, we will test a
> - longword at a time. The tricky part is testing if *any of the four*
> - bytes in the longword in question are equal to c. We first use an xor
> - with repeated_c. This reduces the task to testing whether *any of the
> - four* bytes in longword1 is zero.
> -
> - We compute tmp =
> - ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
> - That is, we perform the following operations:
> - 1. Subtract repeated_one.
> - 2. & ~longword1.
> - 3. & a mask consisting of 0x80 in every byte.
> - Consider what happens in each byte:
> - - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
> - and step 3 transforms it into 0x80. A carry can also be propagated
> - to more significant bytes.
> - - If a byte of longword1 is nonzero, let its lowest 1 bit be at
> - position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
> - the byte ends in a single bit of value 0 and k bits of value 1.
> - After step 2, the result is just k bits of value 1: 2^k - 1. After
> - step 3, the result is 0. And no carry is produced.
> - So, if longword1 has only non-zero bytes, tmp is zero.
> - Whereas if longword1 has a zero byte, call j the position of the least
> - significant zero byte. Then the result has a zero at positions 0, ...,
> - j-1 and a 0x80 at position j. We cannot predict the result at the more
> - significant bytes (positions j+1..3), but it does not matter since we
> - already have a non-zero bit at position 8*j+7.
> -
> - So, the test whether any byte in longword1 is zero is equivalent to
> - testing whether tmp is nonzero. */
> -
> - while (n >= sizeof (longword))
> - {
> - longword longword1 = *longword_ptr ^ repeated_c;
> -
> - if ((((longword1 - repeated_one) & ~longword1)
> - & (repeated_one << 7)) != 0)
> - break;
> - longword_ptr++;
> - n -= sizeof (longword);
> - }
> -
> - char_ptr = (const unsigned char *) longword_ptr;
> -
> - /* At this point, we know that either n < sizeof (longword), or one of the
> - sizeof (longword) bytes starting at char_ptr is == c. On little-endian
> - machines, we could determine the first such byte without any further
> - memory accesses, just by looking at the tmp result from the last loop
> - iteration. But this does not work on big-endian machines. Choose code
> - that works in both cases. */
> -
> - for (; n > 0; --n, ++char_ptr)
> - {
> - if (*char_ptr == c)
> - return (void *) char_ptr;
> - }
> -
> - return NULL;
> + /* We found a match, but it might be in a byte past the end
> + of the array. */
> + char *ret = (char *) word_ptr + index_first_eq (word, repeated_c);
> + return (ret <= lbyte) ? ret : NULL;
> }
> -#ifdef weak_alias
> +#ifndef MEMCHR
> weak_alias (__memchr, memchr)
> -#endif
> libc_hidden_builtin_def (memchr)
> +#endif
> diff --git a/sysdeps/powerpc/powerpc32/power4/multiarch/memchr-ppc32.c b/sysdeps/powerpc/powerpc32/power4/multiarch/memchr-ppc32.c
> index fc69df54b3..02877d3c98 100644
> --- a/sysdeps/powerpc/powerpc32/power4/multiarch/memchr-ppc32.c
> +++ b/sysdeps/powerpc/powerpc32/power4/multiarch/memchr-ppc32.c
> @@ -18,17 +18,11 @@
>
> #include <string.h>
>
> -#define MEMCHR __memchr_ppc
> +extern __typeof (memchr) __memchr_ppc attribute_hidden;
>
> -#undef weak_alias
> -#define weak_alias(a, b)
> +#define MEMCHR __memchr_ppc
> +#include <string/memchr.c>
>
> #ifdef SHARED
> -# undef libc_hidden_builtin_def
> -# define libc_hidden_builtin_def(name) \
> - __hidden_ver1(__memchr_ppc, __GI_memchr, __memchr_ppc);
> +__hidden_ver1(__memchr_ppc, __GI_memchr, __memchr_ppc);
> #endif
> -
> -extern __typeof (memchr) __memchr_ppc attribute_hidden;
> -
> -#include <string/memchr.c>
> diff --git a/sysdeps/powerpc/powerpc64/multiarch/memchr-ppc64.c b/sysdeps/powerpc/powerpc64/multiarch/memchr-ppc64.c
> index 3c966f4403..15beca787b 100644
> --- a/sysdeps/powerpc/powerpc64/multiarch/memchr-ppc64.c
> +++ b/sysdeps/powerpc/powerpc64/multiarch/memchr-ppc64.c
> @@ -18,14 +18,7 @@
>
> #include <string.h>
>
> -#define MEMCHR __memchr_ppc
> -
> -#undef weak_alias
> -#define weak_alias(a, b)
> -
> -# undef libc_hidden_builtin_def
> -# define libc_hidden_builtin_def(name)
> -
> extern __typeof (memchr) __memchr_ppc attribute_hidden;
>
> +#define MEMCHR __memchr_ppc
> #include <string/memchr.c>
> --
> 2.34.1
>
On 03/09/22 00:47, Noah Goldstein wrote:
>>
>> - longword_ptr = (const longword *) char_ptr;
>> + /* Compute the address of the word containing the last byte. */
>> + const op_t *lword = word_containing (lbyte);
>>
>> - /* All these elucidatory comments refer to 4-byte longwords,
>> - but the theory applies equally well to any size longwords. */
>> + /* Read the first word, but munge it so that bytes before the array
>> + will not match goal. */
>> + const op_t * word_ptr = word_containing (s);
>> + op_t word = (*word_ptr | before_mask) ^ (repeated_c & before_mask);
>
> Why do you xor with repeated_c & before_mask here?
>
> Doesn't the has_eq(word, repeated_c) do that?
For the case of c_in being 0xff, since for this case or with before_mask
will make has_eq to return early. The test-memchr does not trigger it,
but test-memccpy does fail without the XOR.
On Mon, Sep 19, 2022 at 12:17 PM Adhemerval Zanella Netto
<adhemerval.zanella@linaro.org> wrote:
>
>
>
> On 03/09/22 00:47, Noah Goldstein wrote:
>
> >>
> >> - longword_ptr = (const longword *) char_ptr;
> >> + /* Compute the address of the word containing the last byte. */
> >> + const op_t *lword = word_containing (lbyte);
> >>
> >> - /* All these elucidatory comments refer to 4-byte longwords,
> >> - but the theory applies equally well to any size longwords. */
> >> + /* Read the first word, but munge it so that bytes before the array
> >> + will not match goal. */
> >> + const op_t * word_ptr = word_containing (s);
> >> + op_t word = (*word_ptr | before_mask) ^ (repeated_c & before_mask);
> >
> > Why do you xor with repeated_c & before_mask here?
> >
> > Doesn't the has_eq(word, repeated_c) do that?
>
> For the case of c_in being 0xff, since for this case or with before_mask
> will make has_eq to return early. The test-memchr does not trigger it,
> but test-memccpy does fail without the XOR.
I see. Since a match in the first several bytes is fairly common
maybe it would be better to special case the first iteration and just do
has_eq(word, repeated_c) >> (CHAR_BIT * (addr % sizeof(addr)).
The result can just be added to `s` if there is a match.
On 19/09/22 18:59, Noah Goldstein wrote:
> On Mon, Sep 19, 2022 at 12:17 PM Adhemerval Zanella Netto
> <adhemerval.zanella@linaro.org> wrote:
>>
>>
>>
>> On 03/09/22 00:47, Noah Goldstein wrote:
>>
>>>>
>>>> - longword_ptr = (const longword *) char_ptr;
>>>> + /* Compute the address of the word containing the last byte. */
>>>> + const op_t *lword = word_containing (lbyte);
>>>>
>>>> - /* All these elucidatory comments refer to 4-byte longwords,
>>>> - but the theory applies equally well to any size longwords. */
>>>> + /* Read the first word, but munge it so that bytes before the array
>>>> + will not match goal. */
>>>> + const op_t * word_ptr = word_containing (s);
>>>> + op_t word = (*word_ptr | before_mask) ^ (repeated_c & before_mask);
>>>
>>> Why do you xor with repeated_c & before_mask here?
>>>
>>> Doesn't the has_eq(word, repeated_c) do that?
>>
>> For the case of c_in being 0xff, since for this case or with before_mask
>> will make has_eq to return early. The test-memchr does not trigger it,
>> but test-memccpy does fail without the XOR.
>
> I see. Since a match in the first several bytes is fairly common
> maybe it would be better to special case the first iteration and just do
>
> has_eq(word, repeated_c) >> (CHAR_BIT * (addr % sizeof(addr)).
> The result can just be added to `s` if there is a match.
I think you mean something like:
has_eq (word >> (CHAR_BIT * (s % sizeof(op_t)), repeated_c)
Since has_eq returns _Bool. However in this case we will need to shift
the repeated_c as well, and it will bleed endianess definition (the shift
direction) on generic implementation. On both cases not sure if this will
be a gain.
Maybe we can also parametrize the first check:
static inline _Bool
has_eq_first (op_t *word, const op_t *word_ptr, op_t repeated_c,
op_t before_mask)
{
*word = (*word_ptr | before_mask) ^ (repeated_c & before_mask);
return has_eq (*word, repeated_c);
}
[...]
op_t word;
if (!has_eq_first (&word, word_ptr, repeated_c, before_mask))
{
do
{
if (word_ptr == lword)
return NULL;
word = *++word_ptr;
}
while (!has_eq (word, repeated_c));
}
If the architecture has a better strategy to check. But I also not sure
if this would indeed yield any improvement in the end.
@@ -1,10 +1,6 @@
-/* Copyright (C) 1991-2022 Free Software Foundation, Inc.
+/* Scan memory for a character. Generic version
+ Copyright (C) 1991-2022 Free Software Foundation, Inc.
This file is part of the GNU C Library.
- Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
- with help from Dan Sahlin (dan@sics.se) and
- commentary by Jim Blandy (jimb@ai.mit.edu);
- adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
- and implemented by Roland McGrath (roland@ai.mit.edu).
The GNU C Library is free software; you can redistribute it and/or
modify it under the terms of the GNU Lesser General Public
@@ -20,143 +16,65 @@
License along with the GNU C Library; if not, see
<https://www.gnu.org/licenses/>. */
-#ifndef _LIBC
-# include <config.h>
-#endif
-
+#include <intprops.h>
+#include <string-fza.h>
+#include <string-fzb.h>
+#include <string-fzi.h>
+#include <string-maskoff.h>
+#include <string-opthr.h>
#include <string.h>
-#include <stddef.h>
+#undef memchr
-#include <limits.h>
-
-#undef __memchr
-#ifdef _LIBC
-# undef memchr
+#ifdef MEMCHR
+# define __memchr MEMCHR
#endif
-#ifndef weak_alias
-# define __memchr memchr
-#endif
-
-#ifndef MEMCHR
-# define MEMCHR __memchr
-#endif
+static inline const char *
+sadd (uintptr_t x, uintptr_t y)
+{
+ uintptr_t ret = INT_ADD_OVERFLOW (x, y) ? (uintptr_t)-1 : x + y;
+ return (const char *)ret;
+}
/* Search no more than N bytes of S for C. */
void *
-MEMCHR (void const *s, int c_in, size_t n)
+__memchr (void const *s, int c_in, size_t n)
{
- /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
- long instead of a 64-bit uintmax_t tends to give better
- performance. On 64-bit hardware, unsigned long is generally 64
- bits already. Change this typedef to experiment with
- performance. */
- typedef unsigned long int longword;
+ if (__glibc_unlikely (n == 0))
+ return NULL;
- const unsigned char *char_ptr;
- const longword *longword_ptr;
- longword repeated_one;
- longword repeated_c;
- unsigned char c;
+ uintptr_t s_int = (uintptr_t) s;
- c = (unsigned char) c_in;
+ /* Set up a word, each of whose bytes is C. */
+ op_t repeated_c = repeat_bytes (c_in);
+ op_t before_mask = create_mask (s_int);
- /* Handle the first few bytes by reading one byte at a time.
- Do this until CHAR_PTR is aligned on a longword boundary. */
- for (char_ptr = (const unsigned char *) s;
- n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
- --n, ++char_ptr)
- if (*char_ptr == c)
- return (void *) char_ptr;
+ /* Compute the address of the last byte taking in consideration possible
+ overflow. */
+ const char *lbyte = sadd (s_int, n - 1);
- longword_ptr = (const longword *) char_ptr;
+ /* Compute the address of the word containing the last byte. */
+ const op_t *lword = word_containing (lbyte);
- /* All these elucidatory comments refer to 4-byte longwords,
- but the theory applies equally well to any size longwords. */
+ /* Read the first word, but munge it so that bytes before the array
+ will not match goal. */
+ const op_t * word_ptr = word_containing (s);
+ op_t word = (*word_ptr | before_mask) ^ (repeated_c & before_mask);
- /* Compute auxiliary longword values:
- repeated_one is a value which has a 1 in every byte.
- repeated_c has c in every byte. */
- repeated_one = 0x01010101;
- repeated_c = c | (c << 8);
- repeated_c |= repeated_c << 16;
- if (0xffffffffU < (longword) -1)
+ while (has_eq (word, repeated_c) == 0)
{
- repeated_one |= repeated_one << 31 << 1;
- repeated_c |= repeated_c << 31 << 1;
- if (8 < sizeof (longword))
- {
- size_t i;
-
- for (i = 64; i < sizeof (longword) * 8; i *= 2)
- {
- repeated_one |= repeated_one << i;
- repeated_c |= repeated_c << i;
- }
- }
+ if (word_ptr == lword)
+ return NULL;
+ word = *++word_ptr;
}
- /* Instead of the traditional loop which tests each byte, we will test a
- longword at a time. The tricky part is testing if *any of the four*
- bytes in the longword in question are equal to c. We first use an xor
- with repeated_c. This reduces the task to testing whether *any of the
- four* bytes in longword1 is zero.
-
- We compute tmp =
- ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
- That is, we perform the following operations:
- 1. Subtract repeated_one.
- 2. & ~longword1.
- 3. & a mask consisting of 0x80 in every byte.
- Consider what happens in each byte:
- - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
- and step 3 transforms it into 0x80. A carry can also be propagated
- to more significant bytes.
- - If a byte of longword1 is nonzero, let its lowest 1 bit be at
- position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
- the byte ends in a single bit of value 0 and k bits of value 1.
- After step 2, the result is just k bits of value 1: 2^k - 1. After
- step 3, the result is 0. And no carry is produced.
- So, if longword1 has only non-zero bytes, tmp is zero.
- Whereas if longword1 has a zero byte, call j the position of the least
- significant zero byte. Then the result has a zero at positions 0, ...,
- j-1 and a 0x80 at position j. We cannot predict the result at the more
- significant bytes (positions j+1..3), but it does not matter since we
- already have a non-zero bit at position 8*j+7.
-
- So, the test whether any byte in longword1 is zero is equivalent to
- testing whether tmp is nonzero. */
-
- while (n >= sizeof (longword))
- {
- longword longword1 = *longword_ptr ^ repeated_c;
-
- if ((((longword1 - repeated_one) & ~longword1)
- & (repeated_one << 7)) != 0)
- break;
- longword_ptr++;
- n -= sizeof (longword);
- }
-
- char_ptr = (const unsigned char *) longword_ptr;
-
- /* At this point, we know that either n < sizeof (longword), or one of the
- sizeof (longword) bytes starting at char_ptr is == c. On little-endian
- machines, we could determine the first such byte without any further
- memory accesses, just by looking at the tmp result from the last loop
- iteration. But this does not work on big-endian machines. Choose code
- that works in both cases. */
-
- for (; n > 0; --n, ++char_ptr)
- {
- if (*char_ptr == c)
- return (void *) char_ptr;
- }
-
- return NULL;
+ /* We found a match, but it might be in a byte past the end
+ of the array. */
+ char *ret = (char *) word_ptr + index_first_eq (word, repeated_c);
+ return (ret <= lbyte) ? ret : NULL;
}
-#ifdef weak_alias
+#ifndef MEMCHR
weak_alias (__memchr, memchr)
-#endif
libc_hidden_builtin_def (memchr)
+#endif
@@ -18,17 +18,11 @@
#include <string.h>
-#define MEMCHR __memchr_ppc
+extern __typeof (memchr) __memchr_ppc attribute_hidden;
-#undef weak_alias
-#define weak_alias(a, b)
+#define MEMCHR __memchr_ppc
+#include <string/memchr.c>
#ifdef SHARED
-# undef libc_hidden_builtin_def
-# define libc_hidden_builtin_def(name) \
- __hidden_ver1(__memchr_ppc, __GI_memchr, __memchr_ppc);
+__hidden_ver1(__memchr_ppc, __GI_memchr, __memchr_ppc);
#endif
-
-extern __typeof (memchr) __memchr_ppc attribute_hidden;
-
-#include <string/memchr.c>
@@ -18,14 +18,7 @@
#include <string.h>
-#define MEMCHR __memchr_ppc
-
-#undef weak_alias
-#define weak_alias(a, b)
-
-# undef libc_hidden_builtin_def
-# define libc_hidden_builtin_def(name)
-
extern __typeof (memchr) __memchr_ppc attribute_hidden;
+#define MEMCHR __memchr_ppc
#include <string/memchr.c>