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- /* Copyright (C) 1991,93,96,97,99,2000,2003 Free Software Foundation, Inc.
- This file is part of the GNU C Library.
- 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).
- 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>
- #include <limits.h>
- #include "memcopy.h"
- #define LONG_MAX_32_BITS 2147483647
- #undef memchr
- /* Search no more than N bytes of S for C. */
- void attribute_hidden *__memchr (const void * 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;
- c = (unsigned char) c_in;
- /* 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 = (const unsigned char *) s;
- n > 0 && ((unsigned long int) char_ptr
- & (sizeof (longword) - 1)) != 0;
- --n, ++char_ptr)
- if (*char_ptr == c)
- return (void *) char_ptr;
- /* 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. */
- if (sizeof (longword) != 4 && sizeof (longword) != 8)
- abort ();
- #if LONG_MAX <= LONG_MAX_32_BITS
- magic_bits = 0x7efefeff;
- #else
- magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
- #endif
- /* Set up a longword, each of whose bytes is C. */
- charmask = c | (c << 8);
- charmask |= charmask << 16;
- #if LONG_MAX > LONG_MAX_32_BITS
- charmask |= charmask << 32;
- #endif
- /* 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 - 1);
- if (cp[0] == c)
- return (void *) cp;
- if (cp[1] == c)
- return (void *) &cp[1];
- if (cp[2] == c)
- return (void *) &cp[2];
- if (cp[3] == c)
- return (void *) &cp[3];
- #if LONG_MAX > 2147483647
- if (cp[4] == c)
- return (void *) &cp[4];
- if (cp[5] == c)
- return (void *) &cp[5];
- if (cp[6] == c)
- return (void *) &cp[6];
- if (cp[7] == c)
- return (void *) &cp[7];
- #endif
- }
- n -= sizeof (longword);
- }
- char_ptr = (const unsigned char *) longword_ptr;
- while (n-- > 0)
- {
- if (*char_ptr == c)
- return (void *) char_ptr;
- else
- ++char_ptr;
- }
- return 0;
- }
- strong_alias(__memchr, memchr)
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