strlen源码分析

/* 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_tstrlen (str)     const char *str;{  const char *char_ptr;  const unsigned long int *longword_ptr;  unsigned long int longword, himagic, lomagic;  /**   * 数据对齐(data alignment)，是指数据所在的内存地址必须是该数据长度的整数倍，   * 这样CPU的存取速度最快。比如在32位的计算机中，一个WORD为4 byte，则WORD数   * 据的起始地址能被4整除的时候CPU的存取效率比较高。CPU的优化规则大概如下：对于n   * 字节(n = 2,4,8...)的元素，它的首地址能被n整除才能获得最好的性能。   */  /**   * 内存对齐解释：(假设这里是32位机器，long类型为4个字节)   * sizeof(longword-1) = 3   * char_ptr是代表的地址，当charptr的地址不是4的整数倍时，也就是charptr的地址%4 为1,2,3，时 &上3   * 一定是不为0的。   */  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 ();  /*一次判断一个word*/  for (;;)    {  /*判断下一个word*/      longword = *longword_ptr++;      /*判断word中是否存在0, 存在0就找出第一个零出现的位置*/      if (((longword - lomagic) & ~longword & himagic) != 0){  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;  //如果是64为机器，继续后面的判断  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)

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