# HG changeset patch # User Bruno Haible # Date 1209202280 -7200 # Node ID e70f61b955b946ea90c721648fbb8c0b2e9ac43d # Parent ba0b9de13414b60ad72906c7ce9075547cf9849e Superficial improvements and comments. diff -r ba0b9de13414 -r e70f61b955b9 ChangeLog --- a/ChangeLog Fri Apr 25 14:52:38 2008 -0600 +++ b/ChangeLog Sat Apr 26 11:31:20 2008 +0200 @@ -1,3 +1,8 @@ +2008-04-26 Bruno Haible + + * lib/memchr2.c (memchr2): Rename local variables. Add explanatory + comments. + 2008-04-25 Eric Blake Use native fstatat on cygwin 1.7.0. diff -r ba0b9de13414 -r e70f61b955b9 lib/memchr2.c --- a/lib/memchr2.c Fri Apr 25 14:52:38 2008 -0600 +++ b/lib/memchr2.c Sat Apr 26 11:31:20 2008 +0200 @@ -46,14 +46,11 @@ const unsigned char *char_ptr; const longword *longword_ptr; - longword longword1; - longword longword2; - longword magic_bits; - longword charmask1; - longword charmask2; + 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; @@ -61,68 +58,108 @@ 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 longword *) char_ptr; - magic_bits = 0x01010101; - charmask1 = c1 | (c1 << 8); - charmask2 = c2 | (c2 << 8); - charmask1 |= charmask1 << 16; - charmask2 |= charmask2 << 16; + /* 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 < TYPE_MAXIMUM (longword)) { - magic_bits |= magic_bits << 31 << 1; - charmask1 |= charmask1 << 31 << 1; - charmask2 |= charmask2 << 31 << 1; - if (8 < sizeof longword1) - for (i = 64; i < sizeof longword1 * 8; i *= 2) - { - magic_bits |= magic_bits << i; - charmask1 |= charmask1 << i; - charmask2 |= charmask2 << i; - } + repeated_one |= repeated_one << 31 << 1; + repeated_c1 |= repeated_c1 << 31 << 1; + repeated_c2 |= repeated_c2 << 31 << 1; + if (8 < sizeof (longword)) + { + int 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 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. + /* 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. - We first use an xor to convert target bytes into a NUL byte, - since the test for a zero byte is more efficient. For all byte - values except 0x00 and 0x80, subtracting 1 from the byte will - leave the most significant bit unchanged. So detecting 0 is - simply a matter of subtracting from all bytes in parallel, and - checking for a most significant bit that changed to 1. */ + 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. - while (n >= sizeof longword1) + Similary, 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)) { - longword1 = *longword_ptr ^ charmask1; - longword2 = *longword_ptr ^ charmask2; + longword longword1 = *longword_ptr ^ repeated_c1; + longword longword2 = *longword_ptr ^ repeated_c2; - if (((((longword1 - magic_bits) & ~longword1) - | ((longword2 - magic_bits) & ~longword2)) - & (magic_bits << 7)) != 0) + if (((((longword1 - repeated_one) & ~longword1) + | ((longword2 - repeated_one) & ~longword2)) + & (repeated_one << 7)) != 0) break; longword_ptr++; - n -= sizeof longword1; + 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 NULL;