1/* Copyright (C) 1991, 1993, 1997, 2000 Free Software Foundation, Inc.
2   This file is part of the GNU C Library.
3   Written by Torbjorn Granlund (tege@sics.se),
4   with help from Dan Sahlin (dan@sics.se);
5   commentary by Jim Blandy (jimb@ai.mit.edu).
6
7   The GNU C Library is free software; you can redistribute it and/or
8   modify it under the terms of the GNU Lesser General Public
9   License as published by the Free Software Foundation; either
10   version 2.1 of the License, or (at your option) any later version.
11
12   The GNU C Library is distributed in the hope that it will be useful,
13   but WITHOUT ANY WARRANTY; without even the implied warranty of
14   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
15   Lesser General Public License for more details.
16
17   You should have received a copy of the GNU Lesser General Public
18   License along with the GNU C Library; if not, write to the Free
19   Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
20   02111-1307 USA.  */
21
22#include <string.h>
23#include <stdlib.h>
24
25#undef strlen
26
27/* Return the length of the null-terminated string STR.  Scan for
28   the null terminator quickly by testing four bytes at a time.  */
29size_t
30strlen (str)
31     const char *str;
32{
33  const char *char_ptr;
34  const unsigned long int *longword_ptr;
35  unsigned long int longword, magic_bits, himagic, lomagic;
36
37  /* Handle the first few characters by reading one character at a time.
38     Do this until CHAR_PTR is aligned on a longword boundary.  */
39  for (char_ptr = str; ((unsigned long int) char_ptr
40			& (sizeof (longword) - 1)) != 0;
41       ++char_ptr)
42    if (*char_ptr == '\0')
43      return char_ptr - str;
44
45  /* All these elucidatory comments refer to 4-byte longwords,
46     but the theory applies equally well to 8-byte longwords.  */
47
48  longword_ptr = (unsigned long int *) char_ptr;
49
50  /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
51     the "holes."  Note that there is a hole just to the left of
52     each byte, with an extra at the end:
53
54     bits:  01111110 11111110 11111110 11111111
55     bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
56
57     The 1-bits make sure that carries propagate to the next 0-bit.
58     The 0-bits provide holes for carries to fall into.  */
59  magic_bits = 0x7efefeffL;
60  himagic = 0x80808080L;
61  lomagic = 0x01010101L;
62  if (sizeof (longword) > 4)
63    {
64      /* 64-bit version of the magic.  */
65      /* Do the shift in two steps to avoid a warning if long has 32 bits.  */
66      magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
67      himagic = ((himagic << 16) << 16) | himagic;
68      lomagic = ((lomagic << 16) << 16) | lomagic;
69    }
70  if (sizeof (longword) > 8)
71    abort ();
72
73  /* Instead of the traditional loop which tests each character,
74     we will test a longword at a time.  The tricky part is testing
75     if *any of the four* bytes in the longword in question are zero.  */
76  for (;;)
77    {
78      /* We tentatively exit the loop if adding MAGIC_BITS to
79	 LONGWORD fails to change any of the hole bits of LONGWORD.
80
81	 1) Is this safe?  Will it catch all the zero bytes?
82	 Suppose there is a byte with all zeros.  Any carry bits
83	 propagating from its left will fall into the hole at its
84	 least significant bit and stop.  Since there will be no
85	 carry from its most significant bit, the LSB of the
86	 byte to the left will be unchanged, and the zero will be
87	 detected.
88
89	 2) Is this worthwhile?  Will it ignore everything except
90	 zero bytes?  Suppose every byte of LONGWORD has a bit set
91	 somewhere.  There will be a carry into bit 8.  If bit 8
92	 is set, this will carry into bit 16.  If bit 8 is clear,
93	 one of bits 9-15 must be set, so there will be a carry
94	 into bit 16.  Similarly, there will be a carry into bit
95	 24.  If one of bits 24-30 is set, there will be a carry
96	 into bit 31, so all of the hole bits will be changed.
97
98	 The one misfire occurs when bits 24-30 are clear and bit
99	 31 is set; in this case, the hole at bit 31 is not
100	 changed.  If we had access to the processor carry flag,
101	 we could close this loophole by putting the fourth hole
102	 at bit 32!
103
104	 So it ignores everything except 128's, when they're aligned
105	 properly.  */
106
107      longword = *longword_ptr++;
108
109      if (
110#if 0
111	  /* Add MAGIC_BITS to LONGWORD.  */
112	  (((longword + magic_bits)
113
114	    /* Set those bits that were unchanged by the addition.  */
115	    ^ ~longword)
116
117	   /* Look at only the hole bits.  If any of the hole bits
118	      are unchanged, most likely one of the bytes was a
119	      zero.  */
120	   & ~magic_bits)
121#else
122	  ((longword - lomagic) & himagic)
123#endif
124	  != 0)
125	{
126	  /* Which of the bytes was the zero?  If none of them were, it was
127	     a misfire; continue the search.  */
128
129	  const char *cp = (const char *) (longword_ptr - 1);
130
131	  if (cp[0] == 0)
132	    return cp - str;
133	  if (cp[1] == 0)
134	    return cp - str + 1;
135	  if (cp[2] == 0)
136	    return cp - str + 2;
137	  if (cp[3] == 0)
138	    return cp - str + 3;
139	  if (sizeof (longword) > 4)
140	    {
141	      if (cp[4] == 0)
142		return cp - str + 4;
143	      if (cp[5] == 0)
144		return cp - str + 5;
145	      if (cp[6] == 0)
146		return cp - str + 6;
147	      if (cp[7] == 0)
148		return cp - str + 7;
149	    }
150	}
151    }
152}
153