1 /* Find the length of STRING, but scan at most MAXLEN characters.
2 Copyright (C) 1991, 1993, 1997, 2000, 2001 Free Software Foundation, Inc.
3 Contributed by Jakub Jelinek <jakub@redhat.com>.
5 Based on strlen written by Torbjorn Granlund (tege@sics.se),
6 with help from Dan Sahlin (dan@sics.se);
7 commentary by Jim Blandy (jimb@ai.mit.edu).
9 The GNU C Library is free software; you can redistribute it and/or
10 modify it under the terms of the GNU Library General Public License as
11 published by the Free Software Foundation; either version 2 of the
12 License, or (at your option) any later version.
14 The GNU C Library is distributed in the hope that it will be useful,
15 but WITHOUT ANY WARRANTY; without even the implied warranty of
16 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
17 Library General Public License for more details.
19 You should have received a copy of the GNU Library General Public
20 License along with the GNU C Library; see the file COPYING.LIB. If not,
21 write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
22 Boston, MA 02111-1307, USA. */
27 /* Find the length of S, but scan at most MAXLEN characters. If no
28 '\0' terminator is found in that many characters, return MAXLEN. */
30 __strnlen (const char *str
, size_t maxlen
)
32 const char *char_ptr
, *end_ptr
= str
+ maxlen
;
33 const unsigned long int *longword_ptr
;
34 unsigned long int longword
, magic_bits
, himagic
, lomagic
;
39 if (__builtin_expect (end_ptr
< str
, 0))
40 end_ptr
= (const char *) ~0UL;
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
= str
; ((unsigned long int) char_ptr
45 & (sizeof (longword
) - 1)) != 0;
47 if (*char_ptr
== '\0')
49 if (char_ptr
> end_ptr
)
51 return char_ptr
- str
;
54 /* All these elucidatory comments refer to 4-byte longwords,
55 but the theory applies equally well to 8-byte longwords. */
57 longword_ptr
= (unsigned long int *) char_ptr
;
59 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
60 the "holes." Note that there is a hole just to the left of
61 each byte, with an extra at the end:
63 bits: 01111110 11111110 11111110 11111111
64 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
66 The 1-bits make sure that carries propagate to the next 0-bit.
67 The 0-bits provide holes for carries to fall into. */
68 magic_bits
= 0x7efefeffL
;
69 himagic
= 0x80808080L
;
70 lomagic
= 0x01010101L
;
71 if (sizeof (longword
) > 4)
73 /* 64-bit version of the magic. */
74 /* Do the shift in two steps to avoid a warning if long has 32 bits. */
75 magic_bits
= ((0x7efefefeL
<< 16) << 16) | 0xfefefeffL
;
76 himagic
= ((himagic
<< 16) << 16) | himagic
;
77 lomagic
= ((lomagic
<< 16) << 16) | lomagic
;
79 if (sizeof (longword
) > 8)
82 /* Instead of the traditional loop which tests each character,
83 we will test a longword at a time. The tricky part is testing
84 if *any of the four* bytes in the longword in question are zero. */
85 while (longword_ptr
< (unsigned long int *) end_ptr
)
87 /* We tentatively exit the loop if adding MAGIC_BITS to
88 LONGWORD fails to change any of the hole bits of LONGWORD.
90 1) Is this safe? Will it catch all the zero bytes?
91 Suppose there is a byte with all zeros. Any carry bits
92 propagating from its left will fall into the hole at its
93 least significant bit and stop. Since there will be no
94 carry from its most significant bit, the LSB of the
95 byte to the left will be unchanged, and the zero will be
98 2) Is this worthwhile? Will it ignore everything except
99 zero bytes? Suppose every byte of LONGWORD has a bit set
100 somewhere. There will be a carry into bit 8. If bit 8
101 is set, this will carry into bit 16. If bit 8 is clear,
102 one of bits 9-15 must be set, so there will be a carry
103 into bit 16. Similarly, there will be a carry into bit
104 24. If one of bits 24-30 is set, there will be a carry
105 into bit 31, so all of the hole bits will be changed.
107 The one misfire occurs when bits 24-30 are clear and bit
108 31 is set; in this case, the hole at bit 31 is not
109 changed. If we had access to the processor carry flag,
110 we could close this loophole by putting the fourth hole
113 So it ignores everything except 128's, when they're aligned
116 longword
= *longword_ptr
++;
118 if ((longword
- lomagic
) & himagic
)
120 /* Which of the bytes was the zero? If none of them were, it was
121 a misfire; continue the search. */
123 const char *cp
= (const char *) (longword_ptr
- 1);
137 if (sizeof (longword
) > 4)
156 if (char_ptr
> end_ptr
)
158 return char_ptr
- str
;
160 weak_alias (__strnlen
, strnlen
)