sysdeps/generic/strlen.c

   1 /* Copyright (C) 1991, 1993, 1997 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
  23 #undef strlen
  24
  25 /* Return the length of the null-terminated string STR.  Scan for
  26    the null terminator quickly by testing four bytes at a time.  */
  27 size_t
  28 strlen (str)
  29      const char *str;
  30 {
  31   const char *char_ptr;
  32   const unsigned long int *longword_ptr;
  33   unsigned long int longword, magic_bits, himagic, lomagic;
  34
  35   /* Handle the first few characters by reading one character at a time.
  36      Do this until CHAR_PTR is aligned on a longword boundary.  */
  37   for (char_ptr = str; ((unsigned long int) char_ptr
  38                         & (sizeof (longword) - 1)) != 0;
  39        ++char_ptr)
  40     if (*char_ptr == '\0')
  41       return char_ptr - str;
  42
  43   /* All these elucidatory comments refer to 4-byte longwords,
  44      but the theory applies equally well to 8-byte longwords.  */
  45
  46   longword_ptr = (unsigned long int *) char_ptr;
  47
  48   /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
  49      the "holes."  Note that there is a hole just to the left of
  50      each byte, with an extra at the end:
  51
  52      bits:  01111110 11111110 11111110 11111111
  53      bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
  54
  55      The 1-bits make sure that carries propagate to the next 0-bit.
  56      The 0-bits provide holes for carries to fall into.  */
  57   magic_bits = 0x7efefeffL;
  58   himagic = 0x80808080L;
  59   lomagic = 0x01010101L;
  60   if (sizeof (longword) > 4)
  61     {
  62       /* 64-bit version of the magic.  */
  63       /* Do the shift in two steps to avoid a warning if long has 32 bits.  */
  64       magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
  65       himagic = ((himagic << 16) << 16) | himagic;
  66       lomagic = ((lomagic << 16) << 16) | lomagic;
  67     }
  68   if (sizeof (longword) > 8)
  69     abort ();
  70
  71   /* Instead of the traditional loop which tests each character,
  72      we will test a longword at a time.  The tricky part is testing
  73      if *any of the four* bytes in the longword in question are zero.  */
  74   for (;;)
  75     {
  76       /* We tentatively exit the loop if adding MAGIC_BITS to
  77          LONGWORD fails to change any of the hole bits of LONGWORD.
  78
  79          1) Is this safe?  Will it catch all the zero bytes?
  80          Suppose there is a byte with all zeros.  Any carry bits
  81          propagating from its left will fall into the hole at its
  82          least significant bit and stop.  Since there will be no
  83          carry from its most significant bit, the LSB of the
  84          byte to the left will be unchanged, and the zero will be
  85          detected.
  86
  87          2) Is this worthwhile?  Will it ignore everything except
  88          zero bytes?  Suppose every byte of LONGWORD has a bit set
  89          somewhere.  There will be a carry into bit 8.  If bit 8
  90          is set, this will carry into bit 16.  If bit 8 is clear,
  91          one of bits 9-15 must be set, so there will be a carry
  92          into bit 16.  Similarly, there will be a carry into bit
  93          24.  If one of bits 24-30 is set, there will be a carry
  94          into bit 31, so all of the hole bits will be changed.
  95
  96          The one misfire occurs when bits 24-30 are clear and bit
  97          31 is set; in this case, the hole at bit 31 is not
  98          changed.  If we had access to the processor carry flag,
  99          we could close this loophole by putting the fourth hole
 100          at bit 32!
 101
 102          So it ignores everything except 128's, when they're aligned
 103          properly.  */
 104
 105       longword = *longword_ptr++;
 106
 107       if (
 108 #if 0
 109           /* Add MAGIC_BITS to LONGWORD.  */
 110           (((longword + magic_bits)
 111
 112             /* Set those bits that were unchanged by the addition.  */
 113             ^ ~longword)
 114
 115            /* Look at only the hole bits.  If any of the hole bits
 116               are unchanged, most likely one of the bytes was a
 117               zero.  */
 118            & ~magic_bits)
 119 #else
 120           ((longword - lomagic) & himagic)
 121 #endif
 122           != 0)
 123         {
 124           /* Which of the bytes was the zero?  If none of them were, it was
 125              a misfire; continue the search.  */
 126
 127           const char *cp = (const char *) (longword_ptr - 1);
 128
 129           if (cp[0] == 0)
 130             return cp - str;
 131           if (cp[1] == 0)
 132             return cp - str + 1;
 133           if (cp[2] == 0)
 134             return cp - str + 2;
 135           if (cp[3] == 0)
 136             return cp - str + 3;
 137           if (sizeof (longword) > 4)
 138             {
 139               if (cp[4] == 0)
 140                 return cp - str + 4;
 141               if (cp[5] == 0)
 142                 return cp - str + 5;
 143               if (cp[6] == 0)
 144                 return cp - str + 6;
 145               if (cp[7] == 0)
 146                 return cp - str + 7;
 147             }
 148         }
 149     }
 150 }