help with string slicing in C

i have a string and i want the first 5 chars to be removed
this is what i have

void shift(char *string) {
    int i;
    for(i=0;i<5;i++) {
        while(1) {
            *string = *(string+1);
            if (*string=='\0') break;
        }
        
    }
}

its not working, i think it maybe because my string isnt terminated properly, but anyways can you see if this function is right?

---------- Post updated 04-03-11 at 12:06 AM ---------- Previous update was 04-02-11 at 11:17 PM ----------

nvm i got it using

void shift(char *string) {
    int i, k;
    char *h=string;
    int j=strlen(string);
    for(k=0;k<5;k++) {
        for(i=0;i<j;i++) {
            *string = *(string+1);
            string++;
        }
        string=h;
        j=strlen(string);
    }
}

This is one way of manyto whack off the first n chars of string:

char *whack(char *src, size_t start)
{
      if(strlen(src) >= start)
      {
          char *p=strdup(src);
          p+=start;
          strcpy(src, p);
          free(p);
      }
      return src;
}

Calling "free(p)" after "p+=start" will corrupt the heap.

My shot:

char *shift( char *str, size_t count )
{
    size_t len;

    if ( NULL == str )
    {
        return( NULL );
    }

    len = strlen( str );
    if ( len > count )
    {
        return( NULL );
    }

    memmove( str, str + cpunt, len - count + 1 );

    return( str );
}

Fast but no error checking added!

void
shift(char *string)
{
    char *ptr = string;

    while (*(ptr+5)) {
       *ptr = *(ptr+5);
       ptr++;
    }
    *ptr = '\0';
}

The dumb+easy way is to use pointer math:

const char *str="0123456789";

printf("Entire string is %s\n", str);
printf("With 5 lopped off the front:  %s\n", str+5);

Superfast, too. One instruction beats copying an entire string.

should be

 if (count > len) 

Thanks,
Gaurav.

---------- Post updated at 03:08 PM ---------- Previous update was at 02:57 PM ----------
Another way ,

 void * shiftbyn( char *str, size_t count )
{
    size_t len = 0;
    char *newstr = NULL;

    if ( NULL == str )
    {
        return NULL;
    }

    len = strlen( str );
    if ( count > len )
    {
        return NULL;
    }

    newstr = malloc (strlen(str) - count);
    strcpy (newstr, str + count);
    free (str); str = NULL;
    return newstr;
}

Thanks,
Gaurav.

wouldnt this work, since strings are pointers?
buf=buf+5;

and whats the difference between declaring a string like
char *buf;
and
char buf[10];

arent they both pointers? But in some cases the way i declare it makes a difference through...

Strings are NOT pointers. Pointer airthmetic (buf = buf + 5) can only be applied if buf is a pointer (char *buf) and not if it is a string (char buf [20] ) . If you want to do pointer airthmetic on a string, get a pointer to point to the base of the string and perform pointer airthmetic on the "pointer".

char buf [20]; //string 

char *pointer_to_buf = buf   // char pointer to the base of 
                                                // string "buf" 
pointer_to_buf += 5;  // gets the pointer to point to the 
                                     // address of buf [4] 

Read K&R (Pointers and Arrays - chapter 5)

Thanks,
Gaurav.