1 /* Copyright (C) 1991,93,94,95,96,97,99,2000 Free Software Foundation, Inc.
2 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
3 with help from Dan Sahlin (dan@sics.se) and
4 bug fix and commentary by Jim Blandy (jimb@ai.mit.edu);
5 adaptation to strchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
6 and implemented by Roland McGrath (roland@ai.mit.edu).
8 The GNU C Library is free software; you can redistribute it and/or
9 modify it under the terms of the GNU Library General Public License as
10 published by the Free Software Foundation; either version 2 of the
11 License, or (at your option) any later version.
13 The GNU C Library is distributed in the hope that it will be useful,
14 but WITHOUT ANY WARRANTY; without even the implied warranty of
15 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
16 Library General Public License for more details.
18 You should have received a copy of the GNU Library General Public
19 License along with the GNU C Library; see the file COPYING.LIB. If not,
20 write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
21 Boston, MA 02111-1307, USA. */
29 /* Find the first occurrence of C in S. */
35 const unsigned char *char_ptr
;
36 const unsigned long int *longword_ptr
;
37 unsigned long int longword
, magic_bits
, charmask
;
40 c
= (unsigned char) c_in
;
42 /* Handle the first few characters by reading one character at a time.
43 Do this until CHAR_PTR is aligned on a longword boundary. */
44 for (char_ptr
= s
; ((unsigned long int) char_ptr
45 & (sizeof (longword
) - 1)) != 0;
48 return (void *) char_ptr
;
49 else if (*char_ptr
== '\0')
52 /* All these elucidatory comments refer to 4-byte longwords,
53 but the theory applies equally well to 8-byte longwords. */
55 longword_ptr
= (unsigned long int *) char_ptr
;
57 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
58 the "holes." Note that there is a hole just to the left of
59 each byte, with an extra at the end:
61 bits: 01111110 11111110 11111110 11111111
62 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
64 The 1-bits make sure that carries propagate to the next 0-bit.
65 The 0-bits provide holes for carries to fall into. */
66 switch (sizeof (longword
))
68 case 4: magic_bits
= 0x7efefeffL
; break;
69 case 8: magic_bits
= ((0x7efefefeL
<< 16) << 16) | 0xfefefeffL
; break;
74 /* Set up a longword, each of whose bytes is C. */
75 charmask
= c
| (c
<< 8);
76 charmask
|= charmask
<< 16;
77 if (sizeof (longword
) > 4)
78 /* Do the shift in two steps to avoid a warning if long has 32 bits. */
79 charmask
|= (charmask
<< 16) << 16;
80 if (sizeof (longword
) > 8)
83 /* Instead of the traditional loop which tests each character,
84 we will test a longword at a time. The tricky part is testing
85 if *any of the four* bytes in the longword in question are zero. */
88 /* We tentatively exit the loop if adding MAGIC_BITS to
89 LONGWORD fails to change any of the hole bits of LONGWORD.
91 1) Is this safe? Will it catch all the zero bytes?
92 Suppose there is a byte with all zeros. Any carry bits
93 propagating from its left will fall into the hole at its
94 least significant bit and stop. Since there will be no
95 carry from its most significant bit, the LSB of the
96 byte to the left will be unchanged, and the zero will be
99 2) Is this worthwhile? Will it ignore everything except
100 zero bytes? Suppose every byte of LONGWORD has a bit set
101 somewhere. There will be a carry into bit 8. If bit 8
102 is set, this will carry into bit 16. If bit 8 is clear,
103 one of bits 9-15 must be set, so there will be a carry
104 into bit 16. Similarly, there will be a carry into bit
105 24. If one of bits 24-30 is set, there will be a carry
106 into bit 31, so all of the hole bits will be changed.
108 The one misfire occurs when bits 24-30 are clear and bit
109 31 is set; in this case, the hole at bit 31 is not
110 changed. If we had access to the processor carry flag,
111 we could close this loophole by putting the fourth hole
114 So it ignores everything except 128's, when they're aligned
117 3) But wait! Aren't we looking for C as well as zero?
118 Good point. So what we do is XOR LONGWORD with a longword,
119 each of whose bytes is C. This turns each byte that is C
122 longword
= *longword_ptr
++;
124 /* Add MAGIC_BITS to LONGWORD. */
125 if ((((longword
+ magic_bits
)
127 /* Set those bits that were unchanged by the addition. */
130 /* Look at only the hole bits. If any of the hole bits
131 are unchanged, most likely one of the bytes was a
133 & ~magic_bits
) != 0 ||
135 /* That caught zeroes. Now test for C. */
136 ((((longword
^ charmask
) + magic_bits
) ^ ~(longword
^ charmask
))
139 /* Which of the bytes was C or zero?
140 If none of them were, it was a misfire; continue the search. */
142 const unsigned char *cp
= (const unsigned char *) (longword_ptr
- 1);
146 else if (*cp
== '\0')
150 else if (*cp
== '\0')
154 else if (*cp
== '\0')
158 else if (*cp
== '\0')
160 if (sizeof (longword
) > 4)
164 else if (*cp
== '\0')
168 else if (*cp
== '\0')
172 else if (*cp
== '\0')
176 else if (*cp
== '\0')
187 weak_alias (strchr
, index
)