strlen源码分析

最新推荐文章于 2025-02-25 23:38:02 发布

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最新推荐文章于 2025-02-25 23:38:02 发布

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分类专栏： C Glibc

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这篇博客详细分析了C标准库中的strlen函数源码，通过数据对齐和内存扫描技术来快速寻找字符串的结束符。文章介绍了如何在不同字长的机器上优化查找过程，并提供了64位机器的适配考虑。

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/* Copyright (C) 1991-2014 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, 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;

  /**
   * 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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