X-Git-Url: http://erislabs.net/gitweb/?a=blobdiff_plain;f=lib%2Fmemchr2.c;h=b8816a5959d94d760585ed482ec1f90fd1a9421a;hb=fc102289543631b711bc0bbd65c0090262b170b9;hp=540ed9fc3b6a74d4e7b6b22568b79f243ec2c54f;hpb=f91b9f973226f2b434ba24e32845f252a3d6e64b;p=gnulib.git diff --git a/lib/memchr2.c b/lib/memchr2.c index 540ed9fc3..b8816a595 100644 --- a/lib/memchr2.c +++ b/lib/memchr2.c @@ -1,5 +1,5 @@ -/* Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2006, - 2008 Free Software Foundation, Inc. +/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2012 + Free Software Foundation, Inc. Based on strlen implementation by Torbjorn Granlund (tege@sics.se), with help from Dan Sahlin (dan@sics.se) and @@ -35,16 +35,20 @@ along with this program. If not, see . */ 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 uintmax_t *longword_ptr; - uintmax_t longword1; - uintmax_t longword2; - uintmax_t magic_bits; - uintmax_t charmask1; - uintmax_t charmask2; + const longword *longword_ptr; + longword repeated_one; + longword repeated_c1; + longword repeated_c2; unsigned char c1; unsigned char c2; - int i; c1 = (unsigned char) c1_in; c2 = (unsigned char) c2_in; @@ -52,143 +56,109 @@ memchr2 (void const *s, int c1_in, int c2_in, size_t n) if (c1 == c2) return memchr (s, c1, n); - /* 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 && (size_t) char_ptr % sizeof longword1 != 0; + 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. */ - longword_ptr = (const uintmax_t *) 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: - - bits: 01111110 11111110 11111110 11111111 - bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD - - The 1-bits make sure that carries propagate to the next 0-bit. - The 0-bits provide holes for carries to fall into. */ - - /* Set MAGIC_BITS to be this pattern of 1 and 0 bits. - Set CHARMASK to be a longword, each of whose bytes is C. */ - - magic_bits = 0xfefefefe; - charmask1 = c1 | (c1 << 8); - charmask2 = c2 | (c2 << 8); - charmask1 |= charmask2 << 16; - charmask1 |= charmask2 << 16; -#if 0xffffffffU < UINTMAX_MAX - magic_bits |= magic_bits << 32; - charmask1 |= charmask1 << 32; - charmask2 |= charmask2 << 32; - if (8 < sizeof longword1) - for (i = 64; i < sizeof longword1 * 8; i *= 2) - { - magic_bits |= magic_bits << i; - charmask1 |= charmask1 << i; - charmask2 |= charmask2 << i; - } -#endif - magic_bits = (UINTMAX_MAX >> 1) & (magic_bits | 1); - - /* 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 longword1) + /* 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. + + Similarly, 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)) { - /* 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. */ - - longword1 = *longword_ptr ^ charmask1; - longword2 = *longword_ptr++ ^ charmask2; - - /* Add MAGIC_BITS to LONGWORD. */ - if ((((longword1 + magic_bits) - - /* Set those bits that were unchanged by the addition. */ - ^ ~longword1) - - /* 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 - || (((longword2 + magic_bits) ^ ~longword2) & ~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] == c1 || cp[0] == c2) - return (void *) cp; - if (cp[1] == c1 || cp[1] == c2) - return (void *) &cp[1]; - if (cp[2] == c1 || cp[2] == c2) - return (void *) &cp[2]; - if (cp[3] == c1 || cp[3] == c2) - return (void *) &cp[3]; - if (4 < sizeof longword1 && (cp[4] == c1 || cp[4] == c2)) - return (void *) &cp[4]; - if (5 < sizeof longword1 && (cp[5] == c1 || cp[5] == c2)) - return (void *) &cp[5]; - if (6 < sizeof longword1 && (cp[6] == c1 || cp[6] == c2)) - return (void *) &cp[6]; - if (7 < sizeof longword1 && (cp[7] == c1 || cp[7] == c2)) - return (void *) &cp[7]; - if (8 < sizeof longword1) - for (i = 8; i < sizeof longword1; i++) - if (cp[i] == c1 || cp[i] == c2) - return (void *) &cp[i]; - } - - n -= sizeof longword1; + 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; - 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 == 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; - ++char_ptr; + return (void *) char_ptr; } - return 0; + return NULL; }