Re: [PATCH] string/memchr.c: Merge from gnulib

[Will Newton <will dot newton at linaro dot org>]

On 26 June 2014 16:24, Will Newton <will.newton@linaro.org> wrote:
> Merge most of the gnulib implementation of memchr. The changes that
> remain are:
>
>  - copyright header
>  - bp-sym.h removed
>  - reg_char removed
>  - allow MEMCHR to be redefined
>  - non-conforming whitespace changes
>
> The merged code fixes a number of -Wundef warnings and also introduces
> an optimized algorithm. I haven't detected any performance difference
> in the new code which I believe is down to the quite specific
> circumstances required to hit it. However the new code is approximately
> half the size of the old code on AArch64 (which uses generic memchr).
>
> ChangeLog:
>
> 2014-06-26  Will Newton  <will.newton@linaro.org>
>
>         * string/memchr.c: Merge from gnulib.
>         [_LIBC]: Remove conditionals.
>         (__ptr_t): Remove define.
>         (LONG_MAX_32_BITS): Likewise.
>         (LONG_MAX): Likewise.
>         (MEMCHR): Use ANSI prototype and optimize algorithm.
> ---
>  string/memchr.c | 233 +++++++++++++++++++++++---------------------------------
>  1 file changed, 94 insertions(+), 139 deletions(-)

Ping?

> diff --git a/string/memchr.c b/string/memchr.c
> index 7408f33..c8e1f9b 100644
> --- a/string/memchr.c
> +++ b/string/memchr.c
> @@ -20,186 +20,141 @@
>     License along with the GNU C Library; if not, see
>     <http://www.gnu.org/licenses/>.  */
>
> -#ifdef HAVE_CONFIG_H
> -#include <config.h>
> +#ifndef _LIBC
> +# include <config.h>
>  #endif
>
> -#undef __ptr_t
> -#define __ptr_t void *
> +#include <string.h>
>
> -#if defined _LIBC
> -# include <string.h>
> -# include <memcopy.h>
> -#endif
> +#include <stddef.h>
>
> -#if HAVE_STDLIB_H || defined _LIBC
> -# include <stdlib.h>
> -#endif
> +#include <limits.h>
>
> -#if HAVE_LIMITS_H || defined _LIBC
> -# include <limits.h>
> +#undef __memchr
> +#ifdef _LIBC
> +# undef memchr
>  #endif
>
> -#define LONG_MAX_32_BITS 2147483647
> -
> -#ifndef LONG_MAX
> -#define LONG_MAX LONG_MAX_32_BITS
> +#ifndef weak_alias
> +# define __memchr memchr
>  #endif
>
> -#include <sys/types.h>
> -
> -#undef memchr
> -#undef __memchr
> -
>  #ifndef MEMCHR
>  # define MEMCHR __memchr
>  #endif
>
>  /* Search no more than N bytes of S for C.  */
> -__ptr_t
> -MEMCHR (s, c_in, n)
> -     const __ptr_t s;
> -     int c_in;
> -     size_t n;
> +void *
> +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;
> +
>    const unsigned char *char_ptr;
> -  const unsigned long int *longword_ptr;
> -  unsigned long int longword, magic_bits, charmask;
> +  const longword *longword_ptr;
> +  longword repeated_one;
> +  longword repeated_c;
>    unsigned char c;
>
>    c = (unsigned char) c_in;
>
> -  /* Handle the first few characters by reading one character at a time.
> +  /* 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 && ((unsigned long int) char_ptr
> -                & (sizeof (longword) - 1)) != 0;
> +       n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
>         --n, ++char_ptr)
>      if (*char_ptr == c)
> -      return (__ptr_t) char_ptr;
> -
> -  /* All these elucidatory comments refer to 4-byte longwords,
> -     but the theory applies equally well to 8-byte longwords.  */
> -
> -  longword_ptr = (unsigned long int *) char_ptr;
> -
> -  /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
> -     the "holes."  Note that there is a hole just to the left of
> -     each byte, with an extra at the end:
> +      return (void *) char_ptr;
>
> -     bits:  01111110 11111110 11111110 11111111
> -     bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
> +  longword_ptr = (const longword *) char_ptr;
>
> -     The 1-bits make sure that carries propagate to the next 0-bit.
> -     The 0-bits provide holes for carries to fall into.  */
> -
> -  if (sizeof (longword) != 4 && sizeof (longword) != 8)
> -    abort ();
> +  /* All these elucidatory comments refer to 4-byte longwords,
> +     but the theory applies equally well to any size longwords.  */
> +
> +  /* 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)
> +    {
> +      repeated_one |= repeated_one << 31 << 1;
> +      repeated_c |= repeated_c << 31 << 1;
> +      if (8 < sizeof (longword))
> +       {
> +         size_t i;
>
> -#if LONG_MAX <= LONG_MAX_32_BITS
> -  magic_bits = 0x7efefeff;
> -#else
> -  magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
> -#endif
> +         for (i = 64; i < sizeof (longword) * 8; i *= 2)
> +           {
> +             repeated_one |= repeated_one << i;
> +             repeated_c |= repeated_c << i;
> +           }
> +       }
> +    }
>
> -  /* Set up a longword, each of whose bytes is C.  */
> -  charmask = c | (c << 8);
> -  charmask |= charmask << 16;
> -#if LONG_MAX > LONG_MAX_32_BITS
> -  charmask |= charmask << 32;
> -#endif
> +  /* 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.  */
>
> -  /* Instead of the traditional loop which tests each character,
> -     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 zero.  */
>    while (n >= sizeof (longword))
>      {
> -      /* We tentatively exit the loop if adding MAGIC_BITS to
> -        LONGWORD fails to change any of the hole bits of LONGWORD.
> -
> -        1) Is this safe?  Will it catch all the zero bytes?
> -        Suppose there is a byte with all zeros.  Any carry bits
> -        propagating from its left will fall into the hole at its
> -        least significant bit and stop.  Since there will be no
> -        carry from its most significant bit, the LSB of the
> -        byte to the left will be unchanged, and the zero will be
> -        detected.
> -
> -        2) Is this worthwhile?  Will it ignore everything except
> -        zero bytes?  Suppose every byte of LONGWORD has a bit set
> -        somewhere.  There will be a carry into bit 8.  If bit 8
> -        is set, this will carry into bit 16.  If bit 8 is clear,
> -        one of bits 9-15 must be set, so there will be a carry
> -        into bit 16.  Similarly, there will be a carry into bit
> -        24.  If one of bits 24-30 is set, there will be a carry
> -        into bit 31, so all of the hole bits will be changed.
> -
> -        The one misfire occurs when bits 24-30 are clear and bit
> -        31 is set; in this case, the hole at bit 31 is not
> -        changed.  If we had access to the processor carry flag,
> -        we could close this loophole by putting the fourth hole
> -        at bit 32!
> -
> -        So it ignores everything except 128's, when they're aligned
> -        properly.
> -
> -        3) But wait!  Aren't we looking for C, not zero?
> -        Good point.  So what we do is XOR LONGWORD with a longword,
> -        each of whose bytes is C.  This turns each byte that is C
> -        into a zero.  */
> -
> -      longword = *longword_ptr++ ^ charmask;
> -
> -      /* Add MAGIC_BITS to LONGWORD.  */
> -      if ((((longword + magic_bits)
> -
> -           /* Set those bits that were unchanged by the addition.  */
> -           ^ ~longword)
> -
> -          /* Look at only the hole bits.  If any of the hole bits
> -             are unchanged, most likely one of the bytes was a
> -             zero.  */
> -          & ~magic_bits) != 0)
> -       {
> -         /* Which of the bytes was C?  If none of them were, it was
> -            a misfire; continue the search.  */
> -
> -         const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
> -
> -         if (cp[0] == c)
> -           return (__ptr_t) cp;
> -         if (cp[1] == c)
> -           return (__ptr_t) &cp[1];
> -         if (cp[2] == c)
> -           return (__ptr_t) &cp[2];
> -         if (cp[3] == c)
> -           return (__ptr_t) &cp[3];
> -#if LONG_MAX > 2147483647
> -         if (cp[4] == c)
> -           return (__ptr_t) &cp[4];
> -         if (cp[5] == c)
> -           return (__ptr_t) &cp[5];
> -         if (cp[6] == c)
> -           return (__ptr_t) &cp[6];
> -         if (cp[7] == c)
> -           return (__ptr_t) &cp[7];
> -#endif
> -       }
> +      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;
>
> -  while (n-- > 0)
> +  /* 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 (__ptr_t) char_ptr;
> -      else
> -       ++char_ptr;
> +       return (void *) char_ptr;
>      }
>
> -  return 0;
> +  return NULL;
>  }
>  #ifdef weak_alias
>  weak_alias (__memchr, memchr)
> --
> 1.9.3
>



-- 
Will Newton
Toolchain Working Group, Linaro