Quick implementation in C, no copyrights from my side, so use it as you wish. But please note that this is a very weak checksum, so don't use it for serious things :) - but that's what you wanted, isn't it?

This returns an 32-bit integer checksum encoded as an string containing its hex value.
If the checksum function doesn't satisfy your needs, you can change the chk += ((int)(str[i]) * (i + 1)); line to something better (f.e. multiplication, addition and bitwise rotating would be much better).

EDIT: Following hughdbrown's advice and one of the answers he linked, I changed the for loop so it doesn't call strlen with every iteration.

#include <stdio.h>

#include <stdlib.h>

#include <string>



char* hextab = 0123456789ABCDEF;



char* encode_int(int i) {

  char* c = (char*)malloc(sizeof(char) * 9);



  for (int j = 0; j < 4; j++) {

    c[(j << 1)] = hextab[((i % 256) >> 4)];

    c[(j << 1) + 1] = hextab[((i % 256) % 16)];



    i = (i >> 8);

  }

  c[8] = 0;



  return c;

}



int checksum(char* str) {

  int i;

  int chk = 0x12345678;



  for (i = 0; str[i] != ''; i++) {

    chk += ((int)(str[i]) * (i + 1));

  }



  return chk;

}



int main() {

  char* str1 = Teststring;

  char* str2 = Teststring2;



  printf(string: %s, checksum string: %sn, str1, encode_int(checksum(str1)));

  printf(string: %s, checksum string: %sn, str2, encode_int(checksum(str2)));



  return 0;

}