126 lines
4.8 KiB
C
126 lines
4.8 KiB
C
/* Searching in a string.
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Copyright (C) 2008-2022 Free Software Foundation, Inc.
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This file is free software: you can redistribute it and/or modify
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it under the terms of the GNU Lesser General Public License as
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published by the Free Software Foundation; either version 2.1 of the
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License, or (at your option) any later version.
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This file 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 Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public License
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along with this program. If not, see <https://www.gnu.org/licenses/>. */
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#include <config.h>
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/* Specification. */
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#include <string.h>
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/* A function definition is only needed if HAVE_RAWMEMCHR is not defined. */
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#if !HAVE_RAWMEMCHR
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# include <limits.h>
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# include <stdalign.h>
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# include <stdint.h>
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# include "verify.h"
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/* Find the first occurrence of C in S. */
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void *
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rawmemchr (const void *s, int c_in)
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{
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/* Change this typedef to experiment with performance. */
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typedef uintptr_t longword;
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/* If you change the "uintptr_t", you should change UINTPTR_WIDTH to match.
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This verifies that the type does not have padding bits. */
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verify (UINTPTR_WIDTH == UCHAR_WIDTH * sizeof (longword));
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const unsigned char *char_ptr;
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unsigned char c = c_in;
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/* Handle the first few bytes by reading one byte 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;
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(uintptr_t) char_ptr % alignof (longword) != 0;
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++char_ptr)
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if (*char_ptr == c)
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return (void *) char_ptr;
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longword const *longword_ptr = s = char_ptr;
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/* Compute auxiliary longword values:
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repeated_one is a value which has a 1 in every byte.
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repeated_c has c in every byte. */
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longword repeated_one = (longword) -1 / UCHAR_MAX;
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longword repeated_c = repeated_one * c;
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longword repeated_hibit = repeated_one * (UCHAR_MAX / 2 + 1);
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/* Instead of the traditional loop which tests each byte, we will
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test a longword at a time. The tricky part is testing if any of
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the bytes in the longword in question are equal to
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c. We first use an xor with repeated_c. This reduces the task
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to testing whether any of the bytes in longword1 is zero.
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(The following comments assume 8-bit bytes, as POSIX requires;
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the code's use of UCHAR_MAX should work even if bytes have more
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than 8 bits.)
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We compute tmp =
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((longword1 - repeated_one) & ~longword1) & (repeated_one * 0x80).
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That is, we perform the following operations:
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1. Subtract repeated_one.
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2. & ~longword1.
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3. & a mask consisting of 0x80 in every byte.
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Consider what happens in each byte:
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- If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
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and step 3 transforms it into 0x80. A carry can also be propagated
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to more significant bytes.
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- If a byte of longword1 is nonzero, let its lowest 1 bit be at
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position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
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the byte ends in a single bit of value 0 and k bits of value 1.
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After step 2, the result is just k bits of value 1: 2^k - 1. After
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step 3, the result is 0. And no carry is produced.
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So, if longword1 has only non-zero bytes, tmp is zero.
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Whereas if longword1 has a zero byte, call j the position of the least
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significant zero byte. Then the result has a zero at positions 0, ...,
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j-1 and a 0x80 at position j. We cannot predict the result at the more
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significant bytes (positions j+1..3), but it does not matter since we
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already have a non-zero bit at position 8*j+7.
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The test whether any byte in longword1 is zero is equivalent
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to testing whether tmp is nonzero.
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This test can read beyond the end of a string, depending on where
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C_IN is encountered. However, this is considered safe since the
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initialization phase ensured that the read will be aligned,
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therefore, the read will not cross page boundaries and will not
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cause a fault. */
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while (1)
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{
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longword longword1 = *longword_ptr ^ repeated_c;
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if ((((longword1 - repeated_one) & ~longword1) & repeated_hibit) != 0)
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break;
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longword_ptr++;
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}
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char_ptr = s = longword_ptr;
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/* At this point, we know that one of the sizeof (longword) bytes
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starting at char_ptr is == c. If we knew endianness, we
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could determine the first such byte without any further memory
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accesses, just by looking at the tmp result from the last loop
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iteration. However, the following simple and portable code does
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not attempt this potential optimization. */
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while (*char_ptr != c)
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char_ptr++;
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return (void *) char_ptr;
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}
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#endif
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