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-rw-r--r--string/strnlen.c161
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diff --git a/string/strnlen.c b/string/strnlen.c
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+/* Find the length of STRING, but scan at most MAXLEN characters.
+ Copyright (C) 1991,1993,1997,2000,2001,2005 Free Software Foundation, Inc.
+ Contributed by Jakub Jelinek <jakub@redhat.com>.
+
+ Based on strlen written by Torbjorn Granlund (tege@sics.se),
+ with help from Dan Sahlin (dan@sics.se);
+ commentary by Jim Blandy (jimb@ai.mit.edu).
+
+ The GNU C Library is free software; you can redistribute it and/or
+ modify it under the terms of the GNU Lesser General Public License as
+ published by the Free Software Foundation; either version 2.1 of the
+ License, or (at your option) any later version.
+
+ The GNU C Library is distributed in the hope that it will be useful,
+ but WITHOUT ANY WARRANTY; without even the implied warranty of
+ MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
+ Lesser General Public License for more details.
+
+ You should have received a copy of the GNU Lesser General Public
+ License along with the GNU C Library; see the file COPYING.LIB. If not,
+ write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
+ Boston, MA 02111-1307, USA. */
+
+#include <string.h>
+#include <stdlib.h>
+
+/* Find the length of S, but scan at most MAXLEN characters. If no
+ '\0' terminator is found in that many characters, return MAXLEN. */
+size_t
+__strnlen (const char *str, size_t maxlen)
+{
+ const char *char_ptr, *end_ptr = str + maxlen;
+ const unsigned long int *longword_ptr;
+ unsigned long int longword, magic_bits, himagic, lomagic;
+
+ if (maxlen == 0)
+ return 0;
+
+ if (__builtin_expect (end_ptr < str, 0))
+ end_ptr = (const char *) ~0UL;
+
+ /* Handle the first few characters by reading one character at a time.
+ Do this until CHAR_PTR is aligned on a longword boundary. */
+ for (char_ptr = str; ((unsigned long int) char_ptr
+ & (sizeof (longword) - 1)) != 0;
+ ++char_ptr)
+ if (*char_ptr == '\0')
+ {
+ if (char_ptr > end_ptr)
+ char_ptr = end_ptr;
+ return char_ptr - str;
+ }
+
+ /* All these elucidatory comments refer to 4-byte longwords,
+ but the theory applies equally well to 8-byte longwords. */
+
+ longword_ptr = (unsigned long int *) char_ptr;
+
+ /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
+ the "holes." Note that there is a hole just to the left of
+ each byte, with an extra at the end:
+
+ bits: 01111110 11111110 11111110 11111111
+ bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
+
+ The 1-bits make sure that carries propagate to the next 0-bit.
+ The 0-bits provide holes for carries to fall into. */
+ magic_bits = 0x7efefeffL;
+ himagic = 0x80808080L;
+ lomagic = 0x01010101L;
+ if (sizeof (longword) > 4)
+ {
+ /* 64-bit version of the magic. */
+ /* Do the shift in two steps to avoid a warning if long has 32 bits. */
+ magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
+ himagic = ((himagic << 16) << 16) | himagic;
+ lomagic = ((lomagic << 16) << 16) | lomagic;
+ }
+ if (sizeof (longword) > 8)
+ abort ();
+
+ /* Instead of the traditional loop which tests each character,
+ we will test a longword at a time. The tricky part is testing
+ if *any of the four* bytes in the longword in question are zero. */
+ while (longword_ptr < (unsigned long int *) end_ptr)
+ {
+ /* We tentatively exit the loop if adding MAGIC_BITS to
+ LONGWORD fails to change any of the hole bits of LONGWORD.
+
+ 1) Is this safe? Will it catch all the zero bytes?
+ Suppose there is a byte with all zeros. Any carry bits
+ propagating from its left will fall into the hole at its
+ least significant bit and stop. Since there will be no
+ carry from its most significant bit, the LSB of the
+ byte to the left will be unchanged, and the zero will be
+ detected.
+
+ 2) Is this worthwhile? Will it ignore everything except
+ zero bytes? Suppose every byte of LONGWORD has a bit set
+ somewhere. There will be a carry into bit 8. If bit 8
+ is set, this will carry into bit 16. If bit 8 is clear,
+ one of bits 9-15 must be set, so there will be a carry
+ into bit 16. Similarly, there will be a carry into bit
+ 24. If one of bits 24-30 is set, there will be a carry
+ into bit 31, so all of the hole bits will be changed.
+
+ The one misfire occurs when bits 24-30 are clear and bit
+ 31 is set; in this case, the hole at bit 31 is not
+ changed. If we had access to the processor carry flag,
+ we could close this loophole by putting the fourth hole
+ at bit 32!
+
+ So it ignores everything except 128's, when they're aligned
+ properly. */
+
+ longword = *longword_ptr++;
+
+ if ((longword - lomagic) & himagic)
+ {
+ /* Which of the bytes was the zero? If none of them were, it was
+ a misfire; continue the search. */
+
+ const char *cp = (const char *) (longword_ptr - 1);
+
+ char_ptr = cp;
+ if (cp[0] == 0)
+ break;
+ char_ptr = cp + 1;
+ if (cp[1] == 0)
+ break;
+ char_ptr = cp + 2;
+ if (cp[2] == 0)
+ break;
+ char_ptr = cp + 3;
+ if (cp[3] == 0)
+ break;
+ if (sizeof (longword) > 4)
+ {
+ char_ptr = cp + 4;
+ if (cp[4] == 0)
+ break;
+ char_ptr = cp + 5;
+ if (cp[5] == 0)
+ break;
+ char_ptr = cp + 6;
+ if (cp[6] == 0)
+ break;
+ char_ptr = cp + 7;
+ if (cp[7] == 0)
+ break;
+ }
+ }
+ char_ptr = end_ptr;
+ }
+
+ if (char_ptr > end_ptr)
+ char_ptr = end_ptr;
+ return char_ptr - str;
+}
+weak_alias (__strnlen, strnlen)
+libc_hidden_def (strnlen)