Remove first few characters from a string

Question

I need to remove the first 3 characters from an array without any libraries. How would I go about doing this? I know that I can use memmove but I'm working on a system without the standard library, also memmove is for pointers. With memmove I can do this:

void chopN(char *str, size_t n)
{
    assert(n != 0 && str != 0);
    size_t len = strlen(str);
    if (n > len)
        return;  // Or: n = len;
    memmove(str, str+n, len - n + 1);
}

But could I remove characters from an array without memmove or any other standard library functions?

does "remove" mean you also want to free the memory? if not (and it's only about output), you''d simply have your pointer point to the new start, i.e. str += 3 — b.buchhold, Nov 10 '16 at 23:29
Note: see no need for `n != 0` in `assert(n != 0 && str != 0)`. — chux - Reinstate Monica, Nov 10 '16 at 23:37

score 2 · Accepted Answer · answered Nov 10 '16 at 23:29

2

As long as you know the string is at least 3 characters long, you can simply use str + 3.

answered Nov 10 '16 at 23:29

Jonathon Reinhart

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1

Potentially dangerous if you don't save the original pointer and need to `free` it later. Doesn't matter if it is on the stack though. – Cody Nov 10 '16 at 23:34
1

@Cody Sure, you could come up with 1000 ways to hurt yourself in C. Strictly speaking this is the best way to accomplish what the OP asked. – Jonathon Reinhart Nov 10 '16 at 23:40

score 2 · Answer 2 · answered Nov 10 '16 at 23:49

Hmmm: 2 simple while loops should do it.

Some untested code to give you an idea.

void chopN(char *str, size_t n) {
  char *dest = str;

  // find beginning watching out for rump `str`
  while (*str && n--) {
    str++;
  }

  // Copy byte by byte
  while (*src) {
    *dest++ = *src++;
  }

  *dest = '\0';
}

Could add a if (n==0) short-cut if desired.

Vlad from Moscow · Answer 3 · 2016-11-10T23:59:39.430

Here is a function that does not use standard C string functions. n can be less then or equal to strlen( s ). Otherwise the function does nothing.

#include <stdio.h>

char * chopN( char *s, size_t n )
{
    char *src = s;

    while ( *src && n ) --n, ++src;

    if ( n == 0 && src != s )
    {
        for ( char *dst = s; (  *dst++ = *src++ ); );
    }

    return s;
}   

int main(void) 
{
    char s[] = "Hello, World";

    puts( s );
    puts( chopN( s, 7 ) );

    return 0;
}

The program output is

Hello, World
World

If you want that in case when n is greater than strlen( s ) all characters were removed then it is enough to substituted the if statement

if ( n == 0 && src != s )

for this one

if ( src != s )

Guilherme Garcia da Rosa · Answer 4 · 2016-11-10T23:58:06.307

0

You don't need to pass the "amount" of characters as a parameter, you can search for '\0':

void chopN(char *str, size_t n){
    char *aux;
    int i=0,j=0;
    while(str[i]!='\0'){
        if(i>n+1){
            aux[j++]=str[i++];
        }else i++;
    }
    aux[j]='\0';
    str = aux;
}

edited Nov 10 '16 at 23:58

answered Nov 10 '16 at 23:34

Guilherme Garcia da Rosa

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score 0 · Answer 5 · answered Nov 10 '16 at 23:46

Simply start at the new start (str+n) and copy to the old start char by char until you reach the end of the string:

char *str1;
for(str1 = str+n; *str1; *str++=*str1++) 
   ;
*str = 0;

If you want something more powerful, you can e.g., steal a memmove implementation from http://git.musl-libc.org/cgit/musl/tree/src/string/memmove.c (it basically does the same thing, except with some performance tweaks and a decision on which way (left/right) the move goes).

Remove first few characters from a string

5 Answers5