diff options
Diffstat (limited to 'string/memchr.c')
| -rw-r--r-- | string/memchr.c | 233 |
1 files changed, 94 insertions, 139 deletions
diff --git a/string/memchr.c b/string/memchr.c index 7408f33b22..c8e1f9b997 100644 --- a/string/memchr.c +++ b/string/memchr.c @@ -20,186 +20,141 @@ License along with the GNU C Library; if not, see <http://www.gnu.org/licenses/>. */ -#ifdef HAVE_CONFIG_H -#include <config.h> +#ifndef _LIBC +# include <config.h> #endif -#undef __ptr_t -#define __ptr_t void * +#include <string.h> -#if defined _LIBC -# include <string.h> -# include <memcopy.h> -#endif +#include <stddef.h> -#if HAVE_STDLIB_H || defined _LIBC -# include <stdlib.h> -#endif +#include <limits.h> -#if HAVE_LIMITS_H || defined _LIBC -# include <limits.h> +#undef __memchr +#ifdef _LIBC +# undef memchr #endif -#define LONG_MAX_32_BITS 2147483647 - -#ifndef LONG_MAX -#define LONG_MAX LONG_MAX_32_BITS +#ifndef weak_alias +# define __memchr memchr #endif -#include <sys/types.h> - -#undef memchr -#undef __memchr - #ifndef MEMCHR # define MEMCHR __memchr #endif /* Search no more than N bytes of S for C. */ -__ptr_t -MEMCHR (s, c_in, n) - const __ptr_t s; - int c_in; - size_t n; +void * +MEMCHR (void const *s, int c_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 unsigned long int *longword_ptr; - unsigned long int longword, magic_bits, charmask; + const longword *longword_ptr; + longword repeated_one; + longword repeated_c; unsigned char c; c = (unsigned char) c_in; - /* Handle the first few characters by reading one character at a time. + /* 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 && ((unsigned long int) char_ptr - & (sizeof (longword) - 1)) != 0; + n > 0 && (size_t) char_ptr % sizeof (longword) != 0; --n, ++char_ptr) if (*char_ptr == c) - return (__ptr_t) char_ptr; - - /* All these elucidatory comments refer to 4-byte longwords, - but the theory applies equally well to 8-byte longwords. */ - - longword_ptr = (unsigned long int *) char_ptr; - - /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits - the "holes." Note that there is a hole just to the left of - each byte, with an extra at the end: + return (void *) char_ptr; - bits: 01111110 11111110 11111110 11111111 - bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD + longword_ptr = (const longword *) char_ptr; - The 1-bits make sure that carries propagate to the next 0-bit. - The 0-bits provide holes for carries to fall into. */ - - if (sizeof (longword) != 4 && sizeof (longword) != 8) - abort (); + /* 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; -#if LONG_MAX <= LONG_MAX_32_BITS - magic_bits = 0x7efefeff; -#else - magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff; -#endif + for (i = 64; i < sizeof (longword) * 8; i *= 2) + { + repeated_one |= repeated_one << i; + repeated_c |= repeated_c << i; + } + } + } - /* Set up a longword, each of whose bytes is C. */ - charmask = c | (c << 8); - charmask |= charmask << 16; -#if LONG_MAX > LONG_MAX_32_BITS - charmask |= charmask << 32; -#endif + /* 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 c. We first use an xor + with repeated_c. This reduces the task to testing whether *any of the + four* bytes in longword1 is zero. + + 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. + + So, the test whether any byte in longword1 is zero is equivalent to + testing whether tmp is nonzero. */ - /* Instead of the traditional loop which tests each character, - 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 zero. */ while (n >= sizeof (longword)) { - /* We tentatively exit the loop if adding MAGIC_BITS to - LONGWORD fails to change any of the hole bits of LONGWORD. - - 1) Is this safe? Will it catch all the zero bytes? - Suppose there is a byte with all zeros. Any carry bits - propagating from its left will fall into the hole at its - least significant bit and stop. Since there will be no - carry from its most significant bit, the LSB of the - byte to the left will be unchanged, and the zero will be - detected. - - 2) Is this worthwhile? Will it ignore everything except - zero bytes? Suppose every byte of LONGWORD has a bit set - somewhere. There will be a carry into bit 8. If bit 8 - is set, this will carry into bit 16. If bit 8 is clear, - one of bits 9-15 must be set, so there will be a carry - into bit 16. Similarly, there will be a carry into bit - 24. If one of bits 24-30 is set, there will be a carry - into bit 31, so all of the hole bits will be changed. - - The one misfire occurs when bits 24-30 are clear and bit - 31 is set; in this case, the hole at bit 31 is not - changed. If we had access to the processor carry flag, - we could close this loophole by putting the fourth hole - at bit 32! - - So it ignores everything except 128's, when they're aligned - properly. - - 3) But wait! Aren't we looking for C, not zero? - Good point. So what we do is XOR LONGWORD with a longword, - each of whose bytes is C. This turns each byte that is C - into a zero. */ - - longword = *longword_ptr++ ^ charmask; - - /* Add MAGIC_BITS to LONGWORD. */ - if ((((longword + magic_bits) - - /* Set those bits that were unchanged by the addition. */ - ^ ~longword) - - /* Look at only the hole bits. If any of the hole bits - are unchanged, most likely one of the bytes was a - zero. */ - & ~magic_bits) != 0) - { - /* Which of the bytes was C? If none of them were, it was - a misfire; continue the search. */ - - const unsigned char *cp = (const unsigned char *) (longword_ptr - 1); - - if (cp[0] == c) - return (__ptr_t) cp; - if (cp[1] == c) - return (__ptr_t) &cp[1]; - if (cp[2] == c) - return (__ptr_t) &cp[2]; - if (cp[3] == c) - return (__ptr_t) &cp[3]; -#if LONG_MAX > 2147483647 - if (cp[4] == c) - return (__ptr_t) &cp[4]; - if (cp[5] == c) - return (__ptr_t) &cp[5]; - if (cp[6] == c) - return (__ptr_t) &cp[6]; - if (cp[7] == c) - return (__ptr_t) &cp[7]; -#endif - } + longword longword1 = *longword_ptr ^ repeated_c; + if ((((longword1 - repeated_one) & ~longword1) + & (repeated_one << 7)) != 0) + break; + longword_ptr++; n -= sizeof (longword); } char_ptr = (const unsigned char *) longword_ptr; - while (n-- > 0) + /* At this point, we know that either n < sizeof (longword), or one of the + sizeof (longword) bytes starting at char_ptr is == 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. */ + + for (; n > 0; --n, ++char_ptr) { if (*char_ptr == c) - return (__ptr_t) char_ptr; - else - ++char_ptr; + return (void *) char_ptr; } - return 0; + return NULL; } #ifdef weak_alias weak_alias (__memchr, memchr) |