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+/* memrchr -- find the last occurrence of a byte in a memory block
+
+ Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2005
+ Free Software Foundation, Inc.
+
+ Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
+ with help from Dan Sahlin (dan@sics.se) and
+ commentary by Jim Blandy (jimb@ai.mit.edu);
+ adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
+ and implemented by Roland McGrath (roland@ai.mit.edu).
+
+ This program is free software; you can redistribute it and/or modify
+ it under the terms of the GNU General Public License as published by
+ the Free Software Foundation; either version 2, or (at your option)
+ any later version.
+
+ This program 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 General Public License for more details.
+
+ You should have received a copy of the GNU General Public License along
+ with this program; if not, write to the Free Software Foundation,
+ Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA. */
+
+#ifdef HAVE_CONFIG_H
+# include <config.h>
+#endif
+
+#if defined _LIBC
+# include <string.h>
+# include <memcopy.h>
+#else
+# include "memrchr.h"
+# define reg_char char
+#endif
+
+#include <limits.h>
+
+#undef __memrchr
+#undef memrchr
+
+#ifndef weak_alias
+# define __memrchr memrchr
+#endif
+
+/* Search no more than N bytes of S for C. */
+void *
+__memrchr (void const *s, int c_in, size_t n)
+{
+ const unsigned char *char_ptr;
+ const unsigned long int *longword_ptr;
+ unsigned long int longword, magic_bits, charmask;
+ unsigned reg_char c;
+ int i;
+
+ c = (unsigned char) c_in;
+
+ /* Handle the last few characters by reading one character at a time.
+ Do this until CHAR_PTR is aligned on a longword boundary. */
+ for (char_ptr = (const unsigned char *) s + n;
+ n > 0 && (size_t) char_ptr % sizeof longword != 0;
+ --n)
+ if (*--char_ptr == c)
+ return (void *) char_ptr;
+
+ /* All these elucidatory comments refer to 4-byte longwords,
+ but the theory applies equally well to any size longwords. */
+
+ longword_ptr = (const 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. */
+
+ /* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
+ Set CHARMASK to be a longword, each of whose bytes is C. */
+
+ magic_bits = 0xfefefefe;
+ charmask = c | (c << 8);
+ charmask |= charmask << 16;
+#if 0xffffffffU < ULONG_MAX
+ magic_bits |= magic_bits << 32;
+ charmask |= charmask << 32;
+ if (8 < sizeof longword)
+ for (i = 64; i < sizeof longword * 8; i *= 2)
+ {
+ magic_bits |= magic_bits << i;
+ charmask |= charmask << i;
+ }
+#endif
+ magic_bits = (ULONG_MAX >> 1) & (magic_bits | 1);
+
+ /* 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 (n >= sizeof longword)
+ {
+ /* 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.
+
+ 3) But wait! Aren't we looking for C, not zero?
+ Good point. So what we do is XOR LONGWORD with a longword,
+ each of whose bytes is C. This turns each byte that is C
+ into a zero. */
+
+ longword = *--longword_ptr ^ charmask;
+
+ /* Add MAGIC_BITS to LONGWORD. */
+ if ((((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) != 0)
+ {
+ /* Which of the bytes was C? If none of them were, it was
+ a misfire; continue the search. */
+
+ const unsigned char *cp = (const unsigned char *) longword_ptr;
+
+ if (8 < sizeof longword)
+ for (i = sizeof longword - 1; 8 <= i; i--)
+ if (cp[i] == c)
+ return (void *) &cp[i];
+ if (7 < sizeof longword && cp[7] == c)
+ return (void *) &cp[7];
+ if (6 < sizeof longword && cp[6] == c)
+ return (void *) &cp[6];
+ if (5 < sizeof longword && cp[5] == c)
+ return (void *) &cp[5];
+ if (4 < sizeof longword && cp[4] == c)
+ return (void *) &cp[4];
+ if (cp[3] == c)
+ return (void *) &cp[3];
+ if (cp[2] == c)
+ return (void *) &cp[2];
+ if (cp[1] == c)
+ return (void *) &cp[1];
+ if (cp[0] == c)
+ return (void *) cp;
+ }
+
+ n -= sizeof longword;
+ }
+
+ char_ptr = (const unsigned char *) longword_ptr;
+
+ while (n-- > 0)
+ {
+ if (*--char_ptr == c)
+ return (void *) char_ptr;
+ }
+
+ return 0;
+}
+#ifdef weak_alias
+weak_alias (__memrchr, memrchr)
+#endif