sysdeps/generic/strlen.c

   1 /* Copyright (C) 1991, 1993 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
  18 not, write to the Free Software Foundation, Inc., 675 Mass Ave,
  19 Cambridge, MA 02139, USA.  */
  20
  21 #include <ansidecl.h>
  22 #include <string.h>
  23
  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
  28 size_t
  29 DEFUN(strlen, (str), 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       magic_bits = (0x7efefefeL << 32) | 0xfefefeffL;
  64       himagic = (himagic << 32) | himagic;
  65       lomagic = (lomagic << 32) | lomagic;
  66     }
  67   if (sizeof (longword) > 8)
  68     abort ();
  69
  70   /* Instead of the traditional loop which tests each character,
  71      we will test a longword at a time.  The tricky part is testing
  72      if *any of the four* bytes in the longword in question are zero.  */
  73   for (;;)
  74     {
  75       /* We tentatively exit the loop if adding MAGIC_BITS to
  76          LONGWORD fails to change any of the hole bits of LONGWORD.
  77
  78          1) Is this safe?  Will it catch all the zero bytes?
  79          Suppose there is a byte with all zeros.  Any carry bits
  80          propagating from its left will fall into the hole at its
  81          least significant bit and stop.  Since there will be no
  82          carry from its most significant bit, the LSB of the
  83          byte to the left will be unchanged, and the zero will be
  84          detected.
  85
  86          2) Is this worthwhile?  Will it ignore everything except
  87          zero bytes?  Suppose every byte of LONGWORD has a bit set
  88          somewhere.  There will be a carry into bit 8.  If bit 8
  89          is set, this will carry into bit 16.  If bit 8 is clear,
  90          one of bits 9-15 must be set, so there will be a carry
  91          into bit 16.  Similarly, there will be a carry into bit
  92          24.  If one of bits 24-30 is set, there will be a carry
  93          into bit 31, so all of the hole bits will be changed.
  94
  95          The one misfire occurs when bits 24-30 are clear and bit
  96          31 is set; in this case, the hole at bit 31 is not
  97          changed.  If we had access to the processor carry flag,
  98          we could close this loophole by putting the fourth hole
  99          at bit 32!
 100
 101          So it ignores everything except 128's, when they're aligned
 102          properly.  */
 103
 104       longword = *longword_ptr++;
 105
 106       if (
 107 #if 0
 108           /* Add MAGIC_BITS to LONGWORD.  */
 109           (((longword + magic_bits)
 110
 111             /* Set those bits that were unchanged by the addition.  */
 112             ^ ~longword)
 113
 114            /* Look at only the hole bits.  If any of the hole bits
 115               are unchanged, most likely one of the bytes was a
 116               zero.  */
 117            & ~magic_bits)
 118 #else
 119           ((longword - lomagic) & himagic)
 120 #endif
 121           != 0)
 122         {
 123           /* Which of the bytes was the zero?  If none of them were, it was
 124              a misfire; continue the search.  */
 125
 126           CONST char *cp = (CONST char *) (longword_ptr - 1);
 127
 128           if (cp[0] == 0)
 129             return cp - str;
 130           if (cp[1] == 0)
 131             return cp - str + 1;
 132           if (cp[2] == 0)
 133             return cp - str + 2;
 134           if (cp[3] == 0)
 135             return cp - str + 3;
 136           if (sizeof (longword) > 4)
 137             {
 138               if (cp[4] == 0)
 139                 return cp - str + 4;
 140               if (cp[5] == 0)
 141                 return cp - str + 5;
 142               if (cp[6] == 0)
 143                 return cp - str + 6;
 144               if (cp[7] == 0)
 145                 return cp - str + 7;
 146             }
 147         }
 148     }
 149 }