glibc 如何找到函数的实现源码

[Crack]

我下载的glibc2.16的源码，想找strlen函数的源码实现怎么都找不到啊？而且string.h文件里面都是空的？是怎么回事啊？应该怎么看glibc的源代码啊？

[Crack]

自己顶一个吧

[arthurzhuyong]

需要工具， 比如sourceinsight, 对于有些宏可能需要全局搜索.

[Crack]

引用 2 楼 arthurzhuyong 的回复:

需要工具， 比如sourceinsight, 对于有些宏可能需要全局搜索.

我用的其他工具搜索的，没有搜索到

[小笨和漂向北方]

glibc-2.19/string/strlen.c: /* Copyright (C) 1991-2014 Free Software Foundation, Inc. This file is part of the GNU C Library. Written by Torbjorn Granlund ([e-mail]), with help from Dan Sahlin ([e-mail]); commentary by Jim Blandy ([e-mail]). 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, see <http://www.gnu.org/licenses/>. */ #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, 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. */ 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. */ 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 (;;) { longword = *longword_ptr++; if (((longword - lomagic) & ~longword & himagic) != 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)