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Diffstat (limited to 'libc/string/generic/rawmemchr.c')
| -rw-r--r-- | libc/string/generic/rawmemchr.c | 160 |
1 files changed, 160 insertions, 0 deletions
diff --git a/libc/string/generic/rawmemchr.c b/libc/string/generic/rawmemchr.c new file mode 100644 index 000000000..0e2ac1c7e --- /dev/null +++ b/libc/string/generic/rawmemchr.c @@ -0,0 +1,160 @@ +/* Copyright (C) 1991,93,96,97,99,2000,2002 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 rawmemchr + +/* Find the first occurrence of C in S. */ +void *rawmemchr (const void * s, int c_in) +{ + 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; + ((unsigned long int) char_ptr & (sizeof (longword) - 1)) != 0; + ++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 (1) + { + /* 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 + } + } +} |