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authorUlrich Drepper <drepper@redhat.com>2005-12-14 15:06:39 +0000
committerUlrich Drepper <drepper@redhat.com>2005-12-14 15:06:39 +0000
commit9d13fb2413921c713f83efe331e8e4d219c62c6b (patch)
tree2d44d7ac45ab2d147eb8361bbff880c365aa8ad5 /string/strlen.c
parentb6ab06cef4670e02756bcdd4d2c33a49369a4346 (diff)
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Moved to csu/errno-loc.c.
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+/* Copyright (C) 1991, 1993, 1997, 2000, 2003 Free Software Foundation, Inc.
+ This file is part of the GNU C Library.
+ 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; 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>
+
+#undef strlen
+
+/* Return the length of the null-terminated string STR. Scan for
+ the null terminator quickly by testing four bytes at a time. */
+size_t
+strlen (str)
+ const char *str;
+{
+ const char *char_ptr;
+ const unsigned long int *longword_ptr;
+ unsigned long int longword, magic_bits, himagic, lomagic;
+
+ /* 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')
+ 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. */
+ for (;;)
+ {
+ /* 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 (
+#if 0
+ /* Add MAGIC_BITS to LONGWORD. */
+ (((longword + magic_bits)
+
+ /* Set those bits that were unchanged by the addition. */
+ ^ ~longword)
+
+ /* Look at only the hole bits. If any of the hole bits
+ are unchanged, most likely one of the bytes was a
+ zero. */
+ & ~magic_bits)
+#else
+ ((longword - lomagic) & himagic)
+#endif
+ != 0)
+ {
+ /* 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);
+
+ if (cp[0] == 0)
+ return cp - str;
+ if (cp[1] == 0)
+ return cp - str + 1;
+ if (cp[2] == 0)
+ return cp - str + 2;
+ if (cp[3] == 0)
+ return cp - str + 3;
+ if (sizeof (longword) > 4)
+ {
+ if (cp[4] == 0)
+ return cp - str + 4;
+ if (cp[5] == 0)
+ return cp - str + 5;
+ if (cp[6] == 0)
+ return cp - str + 6;
+ if (cp[7] == 0)
+ return cp - str + 7;
+ }
+ }
+ }
+}
+libc_hidden_builtin_def (strlen)