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/* memrchr -- find the last occurrence of a byte in a memory block
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Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2005,
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2006, 2007 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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This program is free software; you can redistribute it and/or modify
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it under the terms of the GNU General Public License as published by
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the Free Software Foundation; either version 2, or (at your option)
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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 along
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with this program; if not, write to the Free Software Foundation,
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Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA. */
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# define reg_char char
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# define __memrchr memrchr
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/* Search no more than N bytes of S for C. */
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__memrchr (void const *s, int c_in, size_t n)
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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_in;
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/* Handle the last 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 + n;
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n > 0 && (size_t) char_ptr % sizeof longword != 0;
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return (void *) 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 any size longwords. */
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longword_ptr = (const 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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/* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
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Set CHARMASK to be a longword, each of whose bytes is C. */
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magic_bits = 0xfefefefe;
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charmask = c | (c << 8);
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charmask |= charmask << 16;
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#if 0xffffffffU < ULONG_MAX
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magic_bits |= magic_bits << 32;
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charmask |= charmask << 32;
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if (8 < sizeof longword)
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for (i = 64; i < sizeof longword * 8; i *= 2)
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magic_bits |= magic_bits << i;
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charmask |= charmask << i;
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magic_bits = (ULONG_MAX >> 1) & (magic_bits | 1);
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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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/* 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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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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So it ignores everything except 128's, when they're aligned
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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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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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/* 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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/* 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;
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if (8 < sizeof longword)
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for (i = sizeof longword - 1; 8 <= i; i--)
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return (void *) &cp[i];
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if (7 < sizeof longword && cp[7] == c)
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return (void *) &cp[7];
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if (6 < sizeof longword && cp[6] == c)
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return (void *) &cp[6];
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if (5 < sizeof longword && cp[5] == c)
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return (void *) &cp[5];
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if (4 < sizeof longword && cp[4] == c)
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return (void *) &cp[4];
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return (void *) &cp[3];
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return (void *) &cp[2];
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return (void *) &cp[1];
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n -= sizeof longword;
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char_ptr = (const unsigned char *) longword_ptr;
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if (*--char_ptr == c)
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return (void *) char_ptr;
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weak_alias (__memrchr, memrchr)
added
removed
gnulib/lib/memrchr.c
