sysdeps/generic/strlen.c

   1 /* Copyright (C) 1991, 1993, 1997, 2000 Free Software Foundation, Inc.
   2    Written by Torbjorn Granlund (tege@sics.se),
   3    with help from Dan Sahlin (dan@sics.se);
   4    commentary by Jim Blandy (jimb@ai.mit.edu).
   5
   6    The GNU C Library is free software; you can redistribute it and/or
   7    modify it under the terms of the GNU Library General Public License as
   8    published by the Free Software Foundation; either version 2 of the
   9    License, or (at your option) any later version.
  10
  11    The GNU C Library is distributed in the hope that it will be useful,
  12    but WITHOUT ANY WARRANTY; without even the implied warranty of
  13    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
  14    Library General Public License for more details.
  15
  16    You should have received a copy of the GNU Library General Public
  17    License along with the GNU C Library; see the file COPYING.LIB.  If not,
  18    write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
  19    Boston, MA 02111-1307, USA.  */
  20
  21 #include <string.h>
  22 #include <stdlib.h>
  23
  24 #undef strlen
  25
  26 /* Return the length of the null-terminated string STR.  Scan for
  27    the null terminator quickly by testing four bytes at a time.  */
  28 size_t
  29 strlen (str)
  30      const char *str;
  31 {
  32   const char *char_ptr;
  33   const unsigned long int *longword_ptr;
  34   unsigned long int longword, magic_bits, himagic, lomagic;
  35
  36   /* Handle the first few characters by reading one character at a time.
  37      Do this until CHAR_PTR is aligned on a longword boundary.  */
  38   for (char_ptr = str; ((unsigned long int) char_ptr
  39                         & (sizeof (longword) - 1)) != 0;
  40        ++char_ptr)
  41     if (*char_ptr == '\0')
  42       return char_ptr - str;
  43
  44   /* All these elucidatory comments refer to 4-byte longwords,
  45      but the theory applies equally well to 8-byte longwords.  */
  46
  47   longword_ptr = (unsigned long int *) char_ptr;
  48
  49   /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
  50      the "holes."  Note that there is a hole just to the left of
  51      each byte, with an extra at the end:
  52
  53      bits:  01111110 11111110 11111110 11111111
  54      bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
  55
  56      The 1-bits make sure that carries propagate to the next 0-bit.
  57      The 0-bits provide holes for carries to fall into.  */
  58   magic_bits = 0x7efefeffL;
  59   himagic = 0x80808080L;
  60   lomagic = 0x01010101L;
  61   if (sizeof (longword) > 4)
  62     {
  63       /* 64-bit version of the magic.  */
  64       /* Do the shift in two steps to avoid a warning if long has 32 bits.  */
  65       magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
  66       himagic = ((himagic << 16) << 16) | himagic;
  67       lomagic = ((lomagic << 16) << 16) | lomagic;
  68     }
  69   if (sizeof (longword) > 8)
  70     abort ();
  71
  72   /* Instead of the traditional loop which tests each character,
  73      we will test a longword at a time.  The tricky part is testing
  74      if *any of the four* bytes in the longword in question are zero.  */
  75   for (;;)
  76     {
  77       /* We tentatively exit the loop if adding MAGIC_BITS to
  78          LONGWORD fails to change any of the hole bits of LONGWORD.
  79
  80          1) Is this safe?  Will it catch all the zero bytes?
  81          Suppose there is a byte with all zeros.  Any carry bits
  82          propagating from its left will fall into the hole at its
  83          least significant bit and stop.  Since there will be no
  84          carry from its most significant bit, the LSB of the
  85          byte to the left will be unchanged, and the zero will be
  86          detected.
  87
  88          2) Is this worthwhile?  Will it ignore everything except
  89          zero bytes?  Suppose every byte of LONGWORD has a bit set
  90          somewhere.  There will be a carry into bit 8.  If bit 8
  91          is set, this will carry into bit 16.  If bit 8 is clear,
  92          one of bits 9-15 must be set, so there will be a carry
  93          into bit 16.  Similarly, there will be a carry into bit
  94          24.  If one of bits 24-30 is set, there will be a carry
  95          into bit 31, so all of the hole bits will be changed.
  96
  97          The one misfire occurs when bits 24-30 are clear and bit
  98          31 is set; in this case, the hole at bit 31 is not
  99          changed.  If we had access to the processor carry flag,
 100          we could close this loophole by putting the fourth hole
 101          at bit 32!
 102
 103          So it ignores everything except 128's, when they're aligned
 104          properly.  */
 105
 106       longword = *longword_ptr++;
 107
 108       if (
 109 #if 0
 110           /* Add MAGIC_BITS to LONGWORD.  */
 111           (((longword + magic_bits)
 112
 113             /* Set those bits that were unchanged by the addition.  */
 114             ^ ~longword)
 115
 116            /* Look at only the hole bits.  If any of the hole bits
 117               are unchanged, most likely one of the bytes was a
 118               zero.  */
 119            & ~magic_bits)
 120 #else
 121           ((longword - lomagic) & himagic)
 122 #endif
 123           != 0)
 124         {
 125           /* Which of the bytes was the zero?  If none of them were, it was
 126              a misfire; continue the search.  */
 127
 128           const char *cp = (const char *) (longword_ptr - 1);
 129
 130           if (cp[0] == 0)
 131             return cp - str;
 132           if (cp[1] == 0)
 133             return cp - str + 1;
 134           if (cp[2] == 0)
 135             return cp - str + 2;
 136           if (cp[3] == 0)
 137             return cp - str + 3;
 138           if (sizeof (longword) > 4)
 139             {
 140               if (cp[4] == 0)
 141                 return cp - str + 4;
 142               if (cp[5] == 0)
 143                 return cp - str + 5;
 144               if (cp[6] == 0)
 145                 return cp - str + 6;
 146               if (cp[7] == 0)
 147                 return cp - str + 7;
 148             }
 149         }
 150     }
 151 }