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Diffstat (limited to 'string/strlen.c')
| -rw-r--r-- | string/strlen.c | 153 |
1 files changed, 153 insertions, 0 deletions
diff --git a/string/strlen.c b/string/strlen.c new file mode 100644 index 0000000..9bc9db6 --- /dev/null +++ b/string/strlen.c @@ -0,0 +1,153 @@ +/* Copyright (C) 1991, 1993, 1997, 2000, 2003 Free Software Foundation, Inc. + This file is part of the GNU C Library. + Written by Torbjorn Granlund (tege@sics.se), + with help from Dan Sahlin (dan@sics.se); + commentary by Jim Blandy (jimb@ai.mit.edu). + + The GNU C Library is free software; you can redistribute it and/or + modify it under the terms of the GNU Lesser General Public + License as published by the Free Software Foundation; either + version 2.1 of the License, or (at your option) any later version. + + The GNU C Library is distributed in the hope that it will be useful, + but WITHOUT ANY WARRANTY; without even the implied warranty of + MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU + Lesser General Public License for more details. + + You should have received a copy of the GNU Lesser General Public + License along with the GNU C Library; if not, write to the Free + Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA + 02111-1307 USA. */ + +#include <string.h> +#include <stdlib.h> + +#undef strlen + +/* Return the length of the null-terminated string STR. Scan for + the null terminator quickly by testing four bytes at a time. */ +size_t +strlen (str) + const char *str; +{ + const char *char_ptr; + const unsigned long int *longword_ptr; + unsigned long int longword, magic_bits, himagic, lomagic; + + /* Handle the first few characters by reading one character at a time. + Do this until CHAR_PTR is aligned on a longword boundary. */ + for (char_ptr = str; ((unsigned long int) char_ptr + & (sizeof (longword) - 1)) != 0; + ++char_ptr) + if (*char_ptr == '\0') + return char_ptr - str; + + /* 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: + + 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. */ + magic_bits = 0x7efefeffL; + himagic = 0x80808080L; + lomagic = 0x01010101L; + if (sizeof (longword) > 4) + { + /* 64-bit version of the magic. */ + /* Do the shift in two steps to avoid a warning if long has 32 bits. */ + magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL; + himagic = ((himagic << 16) << 16) | himagic; + lomagic = ((lomagic << 16) << 16) | lomagic; + } + if (sizeof (longword) > 8) + abort (); + + /* 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. */ + for (;;) + { + /* 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. */ + + longword = *longword_ptr++; + + if ( +#if 0 + /* Add MAGIC_BITS to LONGWORD. */ + (((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) +#else + ((longword - lomagic) & himagic) +#endif + != 0) + { + /* Which of the bytes was the zero? If none of them were, it was + a misfire; continue the search. */ + + const char *cp = (const char *) (longword_ptr - 1); + + if (cp[0] == 0) + return cp - str; + if (cp[1] == 0) + return cp - str + 1; + if (cp[2] == 0) + return cp - str + 2; + if (cp[3] == 0) + return cp - str + 3; + if (sizeof (longword) > 4) + { + if (cp[4] == 0) + return cp - str + 4; + if (cp[5] == 0) + return cp - str + 5; + if (cp[6] == 0) + return cp - str + 6; + if (cp[7] == 0) + return cp - str + 7; + } + } + } +} +libc_hidden_builtin_def (strlen) |