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/* Searching in a string.
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Copyright (C) 2003, 2007-2011 Free Software Foundation, Inc.
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This program is free software: you can redistribute it and/or modify
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it under the terms of the GNU General Public License as published by
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the Free Software Foundation; either version 3 of the License, or
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(at your option) any later version.
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This program 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 General Public License for more details.
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You should have received a copy of the GNU General Public License
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along with this program. If not, see <http://www.gnu.org/licenses/>. */
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/* Find the first occurrence of C in S or the final NUL byte. */
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strchrnul (const char *s, int c_in)
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/* On 32-bit hardware, choosing longword to be a 32-bit unsigned
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long instead of a 64-bit uintmax_t tends to give better
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performance. On 64-bit hardware, unsigned long is generally 64
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bits already. Change this typedef to experiment with
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typedef unsigned long int longword;
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const unsigned char *char_ptr;
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const longword *longword_ptr;
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longword repeated_one;
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c = (unsigned char) c_in;
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return rawmemchr (s, 0);
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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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(size_t) char_ptr % sizeof (longword) != 0;
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if (!*char_ptr || *char_ptr == c)
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return (char *) char_ptr;
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longword_ptr = (const longword *) char_ptr;
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/* All these elucidatory comments refer to 4-byte longwords,
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but the theory applies equally well to any size longwords. */
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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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repeated_one = 0x01010101;
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repeated_c = c | (c << 8);
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repeated_c |= repeated_c << 16;
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if (0xffffffffU < (longword) -1)
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repeated_one |= repeated_one << 31 << 1;
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repeated_c |= repeated_c << 31 << 1;
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if (8 < sizeof (longword))
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for (i = 64; i < sizeof (longword) * 8; i *= 2)
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repeated_one |= repeated_one << i;
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repeated_c |= repeated_c << i;
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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 four* bytes in the longword in question are equal to NUL or
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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 four* bytes in longword1 or
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Let's consider longword1. We compute tmp =
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((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
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That is, we perform the following operations:
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1. Subtract repeated_one.
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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 or longword2 is zero is equivalent
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to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine
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this into a single test, whether (tmp1 | tmp2) is nonzero.
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This test can read more than one byte beyond the end of a string,
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depending on where the terminating NUL is encountered. However,
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this is considered safe since the initialization phase ensured
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that the read will be aligned, therefore, the read will not cross
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page boundaries and will not cause a fault. */
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longword longword1 = *longword_ptr ^ repeated_c;
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longword longword2 = *longword_ptr;
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if (((((longword1 - repeated_one) & ~longword1)
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| ((longword2 - repeated_one) & ~longword2))
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& (repeated_one << 7)) != 0)
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char_ptr = (const unsigned char *) 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 == 0 or == c. On little-endian machines,
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we 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. But this does not work on big-endian machines.
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Choose code that works in both cases. */
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char_ptr = (unsigned char *) longword_ptr;
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while (*char_ptr && (*char_ptr != c))
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return (char *) char_ptr;
added
removed
lib/strchrnul.c
