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char* strchr(char*s,charc)
{
while(*s!='\0'&&*s!=c)
{
++s;
}
return*s==c?s:NULL;
}




// strchr.c查找字符串s中首次出现字符c的位置
#include<stdio.h>
#include<string.h>

char* my_strchr(const char *s1, const char *s2)
{
    const char *p = s1;
    const size_t len = strlen (s2);
    for(;(p = strchr (p, *s2))!=0;p++)
    {
        if(strncmp(p,s2,len)==0)
        return(char*)p;
    }
    return(0);
}

int main(void)
{
/*
    char string[17];
    char *ptr, c = 'a';
    strcpy(string, "This is a string");
    ptr = strchr(string, c);
    //ptr = my_strchr(string, c);
    if (ptr)
    {
        *ptr='\0';
        printf("%s\n", string);
        //printf("The character %c is at position: %d,ptr=%s,string=%s\n", c, ptr-string,ptr,string);
    }
    else
        printf("The character was not found\n");
    return 0; 
*/
/*
char *ifName="atm/(0_0_32)",   *pStart,      *pEnd,     p[16];
pStart=strchr(ifName,'(');
pEnd=strchr(ifName,')');
p[0]='\0';
strncat(p,pStart+1,pEnd-pStart-1);
printf("%s,pStart=%d,pEnd=%d\n",p,pStart,pEnd);
*/

/*
   char wanL2PvcName[32];
   char *pStart,*pEnd,*pvcname='\0';
   strncpy(wanL2PvcName, "atm0/(0_0_45)", sizeof(wanL2PvcName)-1);
   if ((pStart = strchr(wanL2PvcName, '(')) && (pEnd = strchr(wanL2PvcName, ')')))
   {
      //now wanL2PvcName is "0_0_32", "0_0_45" etc. 
      wanL2PvcName[0]='\0';
      strncat(wanL2PvcName,pStart+1,pEnd-pStart-1);
      printf("wanL2PvcName=%s\n",wanL2PvcName);
   }
*/

        char *name="ipoe_0_0_32",*pvcname;
        if (name != NULL)
        {
            *name = '\0';
        }
            if(pvcname = strchr(name,'_'))
            {
                if(strcmp(pvcname+1,"0_0_32") == 0)
                    printf("pvcname+1=%s\n\n",pvcname+1);
            }

}
//函数strstr(),可以用于在一个较长的字符串中查找匹配的字符串或字符。
//string strstr(string str,string needle)
//如果找到了needle的准确位置,将从needle的前面返回一个字符串。否则返回false。如果不止一个needle,返回的字符串从出现的第一个needle开始。
//strchr和strstr()类似。
//stristr()不区分大小写,strchr区分大小写
//strrchr()是从最后出现needle的位置开始返回string值。

 

posted on 2014-11-12 02:38  xxxxxxxx1x2xxxxxxx  阅读(298)  评论(0编辑  收藏  举报