GNU LIBC源代码学习之strcmp

比較两个字符串

我的代码块

#include <string.h>


int my_strcmp(const char* s1,const char * s2)
{
 if((s1==NULL)||(s2==NULL))
  return 0;
 while(1)
 {
  if((*s1=='\0')||(*s2=='\0'))
   break;
  if(*s1>*s2)
   return 1;
  if(*s1<*s2)
   return -1;
  s1++;
  s2++;
 }
 //if(*s1==*s2=='\0')
 // return 0;
 if(*s1=='\0'&&*s2=='\0')
  return 0;
 if(*s1>*s2)
  return 1;
 else
  return -1;

}

int main()
{
 char* a="abcdef";
 char* b="abcdef";
 my_strcmp(a,b);
 return 0;
}

这里,上面的方法。不可靠,没有明白的内在逻辑。
看看源代码

/* Copyright (C) 1991-2015 Free Software Foundation, Inc.
   This file is part of the GNU C Library.

   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>

#undef strcmp

/* Compare S1 and S2, returning less than, equal to or
   greater than zero if S1 is lexicographically less than,
   equal to or greater than S2.  */
int
strcmp (const char *p1, const char *p2)
{
  const unsigned char *s1 = (const unsigned char *) p1;
  const unsigned char *s2 = (const unsigned char *) p2;
  unsigned char c1, c2;

  do
    {
      c1 = (unsigned char) *s1++;
      c2 = (unsigned char) *s2++;
      if (c1 == '\0')
return c1 - c2;
    }
  while (c1 == c2);

  return c1 - c2;
}

要比較s1和s2的每个字节的大小,则,两者比較,最简单的分类就是相等和不相等,在相等的时候继续,不相等时。马上计算返回。

至于
const unsigned char s1 = (const unsigned char ) p1;
const unsigned char s2 = (const unsigned char ) p2;
这样转换的妙处,未能体会

posted @ 2016-03-31 12:06  mfrbuaa  阅读(568)  评论(0编辑  收藏  举报