Re: memchr speed

[Bruno Haible]

Eric Blake asked:
> Also, is anyone interested in making gnulib's memchr and strchrnul more 
> efficient by copying the optimizations learned in memchr2?

Checked in like this:

2008-04-26  Eric Blake  <address@hidden>
            Bruno Haible  <address@hidden>

        * lib/memchr.c: Include intprops.h.
        (__memchr): Optimize parallel detection of matching bytes. Rename local
        variables. Add explanatory comments.

*** lib/memchr.c.orig   2008-04-26 12:34:09.000000000 +0200
--- lib/memchr.c        2008-04-26 12:32:30.000000000 +0200
***************
*** 45,50 ****
--- 45,52 ----
  # define BP_SYM(sym) sym
  #endif
  
+ #include "intprops.h"
+ 
  #undef __memchr
  #ifdef _LIBC
  # undef memchr
***************
*** 58,205 ****
  void *
  __memchr (void const *s, int c_in, size_t n)
  {
    const unsigned char *char_ptr;
!   const unsigned long int *longword_ptr;
!   unsigned long int longword, magic_bits, charmask;
    unsigned reg_char c;
-   int i;
  
    c = (unsigned char) c_in;
  
!   /* Handle the first few characters by reading one character 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;
  
    /* All these elucidatory comments refer to 4-byte longwords,
       but the theory applies equally well to any size longwords.  */
  
!   longword_ptr = (const 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:
! 
!      bits:  01111110 11111110 11111110 11111111
!      bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
! 
!      The 1-bits make sure that carries propagate to the next 0-bit.
!      The 0-bits provide holes for carries to fall into.  */
! 
!   /* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
!      Set CHARMASK to be a longword, each of whose bytes is C.  */
! 
!   magic_bits = 0xfefefefe;
!   charmask = c | (c << 8);
!   charmask |= charmask << 16;
! #if 0xffffffffU < ULONG_MAX
!   magic_bits |= magic_bits << 32;
!   charmask |= charmask << 32;
!   if (8 < sizeof longword)
!     for (i = 64; i < sizeof longword * 8; i *= 2)
!       {
!       magic_bits |= magic_bits << i;
!       charmask |= charmask << i;
!       }
! #endif
!   magic_bits = (ULONG_MAX >> 1) & (magic_bits | 1);
! 
!   /* 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 (void *) cp;
!         if (cp[1] == c)
!           return (void *) &cp[1];
!         if (cp[2] == c)
!           return (void *) &cp[2];
!         if (cp[3] == c)
!           return (void *) &cp[3];
!         if (4 < sizeof longword && cp[4] == c)
!           return (void *) &cp[4];
!         if (5 < sizeof longword && cp[5] == c)
!           return (void *) &cp[5];
!         if (6 < sizeof longword && cp[6] == c)
!           return (void *) &cp[6];
!         if (7 < sizeof longword && cp[7] == c)
!           return (void *) &cp[7];
!         if (8 < sizeof longword)
!           for (i = 8; i < sizeof longword; i++)
!             if (cp[i] == c)
!               return (void *) &cp[i];
        }
  
!       n -= sizeof longword;
      }
  
    char_ptr = (const unsigned char *) longword_ptr;
  
!   while (n-- > 0)
      {
        if (*char_ptr == c)
        return (void *) char_ptr;
-       else
-       ++char_ptr;
      }
  
!   return 0;
  }
  #ifdef weak_alias
  weak_alias (__memchr, BP_SYM (memchr))
--- 60,173 ----
  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 longword;
+ 
    const unsigned char *char_ptr;
!   const longword *longword_ptr;
!   longword repeated_one;
!   longword repeated_c;
    unsigned reg_char c;
  
    c = (unsigned char) c_in;
  
!   /* 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;
  
+   longword_ptr = (const longword *) char_ptr;
+ 
    /* 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 < TYPE_MAXIMUM (longword))
      {
!       repeated_one |= repeated_one << 31 << 1;
!       repeated_c |= repeated_c << 31 << 1;
!       if (8 < sizeof (longword))
        {
!         int i;
  
!         for (i = 64; i < sizeof (longword) * 8; i *= 2)
!           {
!             repeated_one |= repeated_one << i;
!             repeated_c |= repeated_c << i;
!           }
        }
+     }
+ 
+   /* 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;
  }
  #ifdef weak_alias
  weak_alias (__memchr, BP_SYM (memchr))
*** modules/memchr.orig 2008-04-26 12:34:09.000000000 +0200
--- modules/memchr      2008-04-26 12:32:02.000000000 +0200
***************
*** 6,11 ****
--- 6,12 ----
  m4/memchr.m4
  
  Depends-on:
+ intprops
  
  configure.ac:
  gl_FUNC_MEMCHR