指针之6.1

/*
 * Find the first occurrence in 'str' of any of the characters in 'chars' and
 * return a poniter to the location. If none are found, or if 'str' or 'chars'
 * NULL pointers, a NULL pointer is returned.
 *
 * 2010-11-20
 * got it.
 */

#define NULL 0


    /* 返回指针的函数 */
char *find_char( char const *str, char const *chars )
{
    char *cp;
    
    /* check arguments for NULL */
    if( str != NULL && chars != NULL )
    {
        for( ; *str != NULL; str++ )
        {
            for( cp = chars; *cp != '\0'; cp++ )
                if( *str == *cp )
                    return str;
        }    
    }
    return NULL;
}