C函数库中的strspn实现

/**copyright@nciaebupt 转载请注明出处*原型：size_t strspn(const char * str1,const char * str2);*用法：#include <string.h>*功能：若strspn()返回的数值为n,*       则代表字符串str1开头连续有n个字符都是属于字符串str2内的字符.*说明：返回字符串str1开头连续包含字符串str2内的字符数目.*使用C函数库中的strspn*/#include <cstdio>#include <cstring>int main(int args,char ** argv){    char str[] = "12999th";    char keys[] = "1234567890";    int i = strspn(str,keys);    printf("the initial string has %d numbers\n",i);    getchar();    return 0;}/**copyright@nciaebupt 转载请注明出处*原型：size_t strspn(const char * str1,const char * str2);*用法：#include <string.h>*功能：若strspn()返回的数值为n,*       则代表字符串str1开头连续有n个字符都是属于字符串str2内的字符.*说明：返回字符串str1开头连续包含字符串str2内的字符数目.*自己实现strspn*/#include <cstdio>#include <cstring>size_t _strspn(const char * string,const char * control){    const char * str = (const char *)string;    const char * ctrl = (const char *)control;    unsigned char map[32];    int count = 0;    /*clear the map*/    memset(map,0,32*sizeof(unsigned char));    //memset(map,0,32);    /*set bits in control map*/    while(*ctrl)    {        map[*ctrl >> 3] |= (0x01 << (*ctrl & 7));        ctrl++;    }    /*count the str's char num in control*/    if(*str)    {        count = 0;        while((map[*str >> 3] & (0x01 << (*str & 7))))        {            count++;            str++;        }        return count;    }    return 0;}int main(int args,char ** argv){    char str[] = "129th";    char keys[] = "1234567890";    int i = _strspn(str,keys);    printf("the initial string has %d numbers\n",i);    getchar();    return 0;}