/* Searching in a string.
- Copyright (C) 2003, 2007 Free Software Foundation, Inc.
+ Copyright (C) 2003, 2007, 2008 Free Software Foundation, Inc.
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
char *
strchrnul (const char *s, int c_in)
{
- char c = c_in;
- while (*s && (*s != c))
- s++;
+ /* 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;
- return (char *) s;
+ const unsigned char *char_ptr;
+ const longword *longword_ptr;
+ longword repeated_one;
+ longword repeated_c;
+ unsigned char c;
+
+ c = (unsigned char) c_in;
+ if (!c)
+ return rawmemchr (s, 0);
+
+ /* 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;
+ (size_t) char_ptr % sizeof (longword) != 0;
+ ++char_ptr)
+ if (!*char_ptr || *char_ptr == c)
+ return (char *) 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_c has c in every byte. */
+ repeated_one = 0x01010101;
+ repeated_c = c | (c << 8);
+ repeated_c |= repeated_c << 16;
+ if (0xffffffffU < (longword) -1)
+ {
+ repeated_one |= repeated_one << 31 << 1;
+ repeated_c |= repeated_c << 31 << 1;
+ if (8 < sizeof (longword))
+ {
+ size_t i;
+
+ for (i = 64; i < sizeof (longword) * 8; i *= 2)
+ {
+ repeated_one |= repeated_one << i;
+ repeated_c |= repeated_c << 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 NUL or
+ c. We first use an xor with repeated_c. This reduces the task
+ to testing whether *any of the four* bytes in longword1 or
+ longword2 is zero.
+
+ Let's consider longword1. We compute tmp =
+ ((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, tmp 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.
+
+ 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.
+
+ This test can read more than one byte beyond the end of a string,
+ depending on where the terminating NUL is encountered. However,
+ this is considered safe since the initialization phase ensured
+ that the read will be aligned, therefore, the read will not cross
+ page boundaries and will not cause a fault. */
+
+ while (1)
+ {
+ longword longword1 = *longword_ptr ^ repeated_c;
+ longword longword2 = *longword_ptr;
+
+ if (((((longword1 - repeated_one) & ~longword1)
+ | ((longword2 - repeated_one) & ~longword2))
+ & (repeated_one << 7)) != 0)
+ break;
+ longword_ptr++;
+ }
+
+ char_ptr = (const unsigned char *) longword_ptr;
+
+ /* At this point, we know that one of the sizeof (longword) bytes
+ starting at char_ptr is == 0 or == c. On little-endian machines,
+ we could determine the first such byte without any further memory
+ accesses, just by looking at the tmp result from the last loop
+ iteration. But this does not work on big-endian machines.
+ Choose code that works in both cases. */
+
+ char_ptr = (unsigned char *) longword_ptr;
+ while (*char_ptr && (*char_ptr != c))
+ char_ptr++;
+ return (char *) char_ptr;
}
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