1 /* Copyright (C) 1991, 1993, 1994, 1995, 1996 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. */
27 /* Find the first occurrence of C in S. */
30 DEFUN(strchr
, (s
, c
), CONST
char *s AND
int c
)
32 CONST
unsigned char *char_ptr
;
33 CONST
unsigned long int *longword_ptr
;
34 unsigned long int longword
, magic_bits
, charmask
;
36 c
= (unsigned char) c
;
38 /* Handle the first few characters by reading one character at a time.
39 Do this until CHAR_PTR is aligned on a longword boundary. */
40 for (char_ptr
= s
; ((unsigned long int) char_ptr
41 & (sizeof (longword
) - 1)) != 0;
44 return (PTR
) char_ptr
;
45 else if (*char_ptr
== '\0')
48 /* All these elucidatory comments refer to 4-byte longwords,
49 but the theory applies equally well to 8-byte longwords. */
51 longword_ptr
= (unsigned long int *) char_ptr
;
53 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
54 the "holes." Note that there is a hole just to the left of
55 each byte, with an extra at the end:
57 bits: 01111110 11111110 11111110 11111111
58 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
60 The 1-bits make sure that carries propagate to the next 0-bit.
61 The 0-bits provide holes for carries to fall into. */
62 switch (sizeof (longword
))
64 case 4: magic_bits
= 0x7efefeffL
; break;
65 case 8: magic_bits
= (0x7efefefeL
<< 32) | 0xfefefeffL
; break;
70 /* Set up a longword, each of whose bytes is C. */
71 charmask
= c
| (c
<< 8);
72 charmask
|= charmask
<< 16;
73 if (sizeof (longword
) > 4)
74 charmask
|= charmask
<< 32;
75 if (sizeof (longword
) > 8)
78 /* Instead of the traditional loop which tests each character,
79 we will test a longword at a time. The tricky part is testing
80 if *any of the four* bytes in the longword in question are zero. */
83 /* We tentatively exit the loop if adding MAGIC_BITS to
84 LONGWORD fails to change any of the hole bits of LONGWORD.
86 1) Is this safe? Will it catch all the zero bytes?
87 Suppose there is a byte with all zeros. Any carry bits
88 propagating from its left will fall into the hole at its
89 least significant bit and stop. Since there will be no
90 carry from its most significant bit, the LSB of the
91 byte to the left will be unchanged, and the zero will be
94 2) Is this worthwhile? Will it ignore everything except
95 zero bytes? Suppose every byte of LONGWORD has a bit set
96 somewhere. There will be a carry into bit 8. If bit 8
97 is set, this will carry into bit 16. If bit 8 is clear,
98 one of bits 9-15 must be set, so there will be a carry
99 into bit 16. Similarly, there will be a carry into bit
100 24. If one of bits 24-30 is set, there will be a carry
101 into bit 31, so all of the hole bits will be changed.
103 The one misfire occurs when bits 24-30 are clear and bit
104 31 is set; in this case, the hole at bit 31 is not
105 changed. If we had access to the processor carry flag,
106 we could close this loophole by putting the fourth hole
109 So it ignores everything except 128's, when they're aligned
112 3) But wait! Aren't we looking for C as well as zero?
113 Good point. So what we do is XOR LONGWORD with a longword,
114 each of whose bytes is C. This turns each byte that is C
117 longword
= *longword_ptr
++;
119 /* Add MAGIC_BITS to LONGWORD. */
120 if ((((longword
+ magic_bits
)
122 /* Set those bits that were unchanged by the addition. */
125 /* Look at only the hole bits. If any of the hole bits
126 are unchanged, most likely one of the bytes was a
128 & ~magic_bits
) != 0 ||
130 /* That caught zeroes. Now test for C. */
131 ((((longword
^ charmask
) + magic_bits
) ^ ~(longword
^ charmask
))
134 /* Which of the bytes was C or zero?
135 If none of them were, it was a misfire; continue the search. */
137 CONST
unsigned char *cp
= (CONST
unsigned char *) (longword_ptr
- 1);
141 else if (*cp
== '\0')
145 else if (*cp
== '\0')
149 else if (*cp
== '\0')
153 else if (*cp
== '\0')
155 if (sizeof (longword
) > 4)
159 else if (*cp
== '\0')
163 else if (*cp
== '\0')
167 else if (*cp
== '\0')
171 else if (*cp
== '\0')
182 weak_alias (strchr
, index
)