200 lines
6.0 KiB
C
200 lines
6.0 KiB
C
/* Copyright (C) 1991, 1993, 1996 Free Software Foundation, Inc.
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Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
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with help from Dan Sahlin (dan@sics.se) and
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commentary by Jim Blandy (jimb@ai.mit.edu);
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adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
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and implemented by Roland McGrath (roland@ai.mit.edu).
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NOTE: The canonical source of this file is maintained with the GNU C Library.
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Bugs can be reported to bug-glibc@prep.ai.mit.edu.
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This program is free software; you can redistribute it and/or modify it
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under the terms of the GNU General Public License as published by the
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Free Software Foundation; either version 2, or (at your option) any
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later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU General Public License for more details.
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You should have received a copy of the GNU General Public License
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along with this program; if not, write to the Free Software
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Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307,
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USA. */
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#ifdef HAVE_CONFIG_H
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#include <config.h>
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#endif
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#undef __ptr_t
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#if defined (__cplusplus) || (defined (__STDC__) && __STDC__)
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# define __ptr_t void *
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#else /* Not C++ or ANSI C. */
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# define __ptr_t char *
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#endif /* C++ or ANSI C. */
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#if defined (_LIBC)
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# include <string.h>
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#endif
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#if defined (HAVE_LIMITS_H) || defined (_LIBC)
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# include <limits.h>
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#endif
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#define LONG_MAX_32_BITS 2147483647
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#ifndef LONG_MAX
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#define LONG_MAX LONG_MAX_32_BITS
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#endif
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#include <sys/types.h>
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/* Search no more than N bytes of S for C. */
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__ptr_t
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memchr (s, c, n)
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const __ptr_t s;
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int c;
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size_t n;
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{
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const unsigned char *char_ptr;
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const unsigned long int *longword_ptr;
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unsigned long int longword, magic_bits, charmask;
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c = (unsigned char) c;
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/* Handle the first few characters by reading one character at a time.
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Do this until CHAR_PTR is aligned on a longword boundary. */
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for (char_ptr = (const unsigned char *) s;
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n > 0 && ((unsigned long int) char_ptr
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& (sizeof (longword) - 1)) != 0;
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--n, ++char_ptr)
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if (*char_ptr == c)
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return (__ptr_t) char_ptr;
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/* All these elucidatory comments refer to 4-byte longwords,
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but the theory applies equally well to 8-byte longwords. */
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longword_ptr = (unsigned long int *) char_ptr;
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/* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
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the "holes." Note that there is a hole just to the left of
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each byte, with an extra at the end:
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bits: 01111110 11111110 11111110 11111111
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bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
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The 1-bits make sure that carries propagate to the next 0-bit.
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The 0-bits provide holes for carries to fall into. */
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if (sizeof (longword) != 4 && sizeof (longword) != 8)
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abort ();
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#if LONG_MAX <= LONG_MAX_32_BITS
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magic_bits = 0x7efefeff;
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#else
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magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
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#endif
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/* Set up a longword, each of whose bytes is C. */
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charmask = c | (c << 8);
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charmask |= charmask << 16;
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#if LONG_MAX > LONG_MAX_32_BITS
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charmask |= charmask << 32;
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#endif
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/* Instead of the traditional loop which tests each character,
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we will test a longword at a time. The tricky part is testing
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if *any of the four* bytes in the longword in question are zero. */
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while (n >= sizeof (longword))
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{
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/* We tentatively exit the loop if adding MAGIC_BITS to
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LONGWORD fails to change any of the hole bits of LONGWORD.
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1) Is this safe? Will it catch all the zero bytes?
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Suppose there is a byte with all zeros. Any carry bits
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propagating from its left will fall into the hole at its
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least significant bit and stop. Since there will be no
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carry from its most significant bit, the LSB of the
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byte to the left will be unchanged, and the zero will be
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detected.
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2) Is this worthwhile? Will it ignore everything except
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zero bytes? Suppose every byte of LONGWORD has a bit set
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somewhere. There will be a carry into bit 8. If bit 8
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is set, this will carry into bit 16. If bit 8 is clear,
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one of bits 9-15 must be set, so there will be a carry
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into bit 16. Similarly, there will be a carry into bit
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24. If one of bits 24-30 is set, there will be a carry
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into bit 31, so all of the hole bits will be changed.
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The one misfire occurs when bits 24-30 are clear and bit
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31 is set; in this case, the hole at bit 31 is not
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changed. If we had access to the processor carry flag,
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we could close this loophole by putting the fourth hole
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at bit 32!
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So it ignores everything except 128's, when they're aligned
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properly.
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3) But wait! Aren't we looking for C, not zero?
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Good point. So what we do is XOR LONGWORD with a longword,
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each of whose bytes is C. This turns each byte that is C
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into a zero. */
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longword = *longword_ptr++ ^ charmask;
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/* Add MAGIC_BITS to LONGWORD. */
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if ((((longword + magic_bits)
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/* Set those bits that were unchanged by the addition. */
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^ ~longword)
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/* Look at only the hole bits. If any of the hole bits
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are unchanged, most likely one of the bytes was a
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zero. */
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& ~magic_bits) != 0)
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{
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/* Which of the bytes was C? If none of them were, it was
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a misfire; continue the search. */
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const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
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if (cp[0] == c)
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return (__ptr_t) cp;
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if (cp[1] == c)
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return (__ptr_t) &cp[1];
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if (cp[2] == c)
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return (__ptr_t) &cp[2];
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if (cp[3] == c)
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return (__ptr_t) &cp[3];
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#if LONG_MAX > 2147483647
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if (cp[4] == c)
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return (__ptr_t) &cp[4];
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if (cp[5] == c)
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return (__ptr_t) &cp[5];
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if (cp[6] == c)
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return (__ptr_t) &cp[6];
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if (cp[7] == c)
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return (__ptr_t) &cp[7];
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#endif
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}
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n -= sizeof (longword);
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}
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char_ptr = (const unsigned char *) longword_ptr;
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while (n-- > 0)
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{
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if (*char_ptr == c)
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return (__ptr_t) char_ptr;
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else
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++char_ptr;
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}
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return 0;
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}
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