Question

I need a quick checksum (as fast as possilbe) for small strings (20-500 chars).

I need the source code and that must be small! (about 100 LOC max)

If it could generate strings in Base32/64. (or something similar) it would be perfect. Basically the checksums cannot use any “bad” chars.. you know.. the usual (){}[].,;:/+-\| etc

Clarifications

It could be strong/weak, that really doesn’t matter since it is only for behind-the-scenes purposes.

It need not contain all the data of the original string since I will be only doing comparison with generated checksums, I don’t expect any sort of “decryption”.

Answer 1

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] != '\0'; i++) {
    chk += ((int)(str[i]) * (i + 1));
  }

  return chk;
}

int main() {
  char* str1 = "Teststring";
  char* str2 = "Teststring2";

  printf("string: %s, checksum string: %s\n", str1, encode_int(checksum(str1)));
  printf("string: %s, checksum string: %s\n", str2, encode_int(checksum(str2)));

  return 0;
}