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Diffstat (limited to 'lib/memchr2.c')
| -rw-r--r-- | lib/memchr2.c | 169 |
1 files changed, 169 insertions, 0 deletions
diff --git a/lib/memchr2.c b/lib/memchr2.c new file mode 100644 index 0000000..1ec53db --- /dev/null +++ b/lib/memchr2.c @@ -0,0 +1,169 @@ +/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2023 + 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 in glibc by Roland McGrath (roland@ai.mit.edu). + Extension to memchr2 implemented by Eric Blake (ebb9@byu.net). + + This file 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. + + This file 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 this program. If not, see <https://www.gnu.org/licenses/>. */ + +#include <config.h> + +#include "memchr2.h" + +#include <limits.h> +#include <stdint.h> +#include <string.h> + +/* Return the first address of either C1 or C2 (treated as unsigned + char) that occurs within N bytes of the memory region S. If + neither byte appears, return NULL. */ +void * +memchr2 (void const *s, int c1_in, int c2_in, size_t n) +{ + /* On 32-bit hardware, choosing longword to be a 32-bit unsigned + long instead of a 64-bit uintmax_t tends to give better + performance. On 64-bit hardware, unsigned long is generally 64 + bits already. Change this typedef to experiment with + performance. */ + typedef unsigned long int longword; + + const unsigned char *char_ptr; + void const *void_ptr; + const longword *longword_ptr; + longword repeated_one; + longword repeated_c1; + longword repeated_c2; + unsigned char c1; + unsigned char c2; + + c1 = (unsigned char) c1_in; + c2 = (unsigned char) c2_in; + + if (c1 == c2) + return memchr (s, c1, n); + + /* Handle the first few bytes by reading one byte at a time. + Do this until VOID_PTR is aligned on a longword boundary. */ + for (void_ptr = s; + n > 0 && (uintptr_t) void_ptr % sizeof (longword) != 0; + --n) + { + char_ptr = void_ptr; + if (*char_ptr == c1 || *char_ptr == c2) + return (void *) void_ptr; + void_ptr = char_ptr + 1; + } + + longword_ptr = void_ptr; + + /* All these elucidatory comments refer to 4-byte longwords, + but the theory applies equally well to any size longwords. */ + + /* Compute auxiliary longword values: + repeated_one is a value which has a 1 in every byte. + repeated_c1 has c1 in every byte. + repeated_c2 has c2 in every byte. */ + repeated_one = 0x01010101; + repeated_c1 = c1 | (c1 << 8); + repeated_c2 = c2 | (c2 << 8); + repeated_c1 |= repeated_c1 << 16; + repeated_c2 |= repeated_c2 << 16; + if (0xffffffffU < (longword) -1) + { + repeated_one |= repeated_one << 31 << 1; + repeated_c1 |= repeated_c1 << 31 << 1; + repeated_c2 |= repeated_c2 << 31 << 1; + if (8 < sizeof (longword)) + { + size_t i; + + for (i = 64; i < sizeof (longword) * 8; i *= 2) + { + repeated_one |= repeated_one << i; + repeated_c1 |= repeated_c1 << i; + repeated_c2 |= repeated_c2 << i; + } + } + } + + /* Instead of the traditional loop which tests each byte, 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 equal to c1 or c2. We first use + an xor with repeated_c1 and repeated_c2, respectively. This reduces + the task to testing whether *any of the four* bytes in longword1 or + longword2 is zero. + + Let's consider longword1. We compute tmp1 = + ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7). + That is, we perform the following operations: + 1. Subtract repeated_one. + 2. & ~longword1. + 3. & a mask consisting of 0x80 in every byte. + Consider what happens in each byte: + - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff, + and step 3 transforms it into 0x80. A carry can also be propagated + to more significant bytes. + - If a byte of longword1 is nonzero, let its lowest 1 bit be at + position k (0 <= k <= 7); so the lowest k bits are 0. After step 1, + the byte ends in a single bit of value 0 and k bits of value 1. + After step 2, the result is just k bits of value 1: 2^k - 1. After + step 3, the result is 0. And no carry is produced. + So, if longword1 has only non-zero bytes, tmp1 is zero. + Whereas if longword1 has a zero byte, call j the position of the least + significant zero byte. Then the result has a zero at positions 0, ..., + j-1 and a 0x80 at position j. We cannot predict the result at the more + significant bytes (positions j+1..3), but it does not matter since we + already have a non-zero bit at position 8*j+7. + + Similarly, we compute tmp2 = + ((longword2 - repeated_one) & ~longword2) & (repeated_one << 7). + + The test whether any byte in longword1 or longword2 is zero is equivalent + to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine + this into a single test, whether (tmp1 | tmp2) is nonzero. */ + + while (n >= sizeof (longword)) + { + longword longword1 = *longword_ptr ^ repeated_c1; + longword longword2 = *longword_ptr ^ repeated_c2; + + if (((((longword1 - repeated_one) & ~longword1) + | ((longword2 - repeated_one) & ~longword2)) + & (repeated_one << 7)) != 0) + break; + longword_ptr++; + n -= sizeof (longword); + } + + char_ptr = (const unsigned char *) longword_ptr; + + /* At this point, we know that either n < sizeof (longword), or one of the + sizeof (longword) bytes starting at char_ptr is == c1 or == c2. On + little-endian machines, we could determine the first such byte without + any further memory accesses, just by looking at the (tmp1 | tmp2) result + from the last loop iteration. But this does not work on big-endian + machines. Choose code that works in both cases. */ + + for (; n > 0; --n, ++char_ptr) + { + if (*char_ptr == c1 || *char_ptr == c2) + return (void *) char_ptr; + } + + return NULL; +} |