/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2011 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 program is free software: you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 3 of the License, or any later version. This program 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 General Public License for more details. You should have received a copy of the GNU General Public License along with this program. If not, see <http://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; 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 CHAR_PTR is aligned on a longword boundary. */ for (char_ptr = (const unsigned char *) s; n > 0 && (size_t) char_ptr % sizeof (longword) != 0; --n, ++char_ptr) if (*char_ptr == c1 || *char_ptr == c2) return (void *) char_ptr; longword_ptr = (const longword *) char_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. Similary, 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; }

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