Re: Convert an ip address to long value

Michael Mair wrote: 
kyle.tk wrote:

I am trying to write a function to convert an ipv4 address that is held
in the string char *ip to its long value equivalent. Here is what I
have right now, but I can't seem to get it to work.

#include <string.h>
#include <stdio.h>

/* Convert an ipv4 address to long integer */
/* "192.168.1.1" --> 3232235777 */
unsigned long iptol(char *ip){
    unsigned char o1,o2,o3,o4; /* The 4 ocets */
    char tmp[13] = "000000000000\0";
    short i = 11; /* Current Index in tmp */
    short j = (strlen(ip) - 1);
    do {
        if ((ip[--j] == '.')){
            i -= (i % 3);
        }
        else {
            tmp[--i] = ip[j];
        }
    } while (i > -1);
    o1 = (tmp[0] * 100) + (tmp[1] * 10) + tmp[2];
    o2 = (tmp[3] * 100) + (tmp[4] * 10) + tmp[5];
    o3 = (tmp[6] * 100) + (tmp[7] * 10) + tmp[8];
    o4 = (tmp[9] * 100) + (tmp[10] * 10) + tmp[11];
    return (o1 * 16777216) + (o2 * 65536) + (o3 * 256) + o4;
}

int main(void){
    char *ip = "124.15.22.102";
    iptol(ip);
}

Any suggestions? Maybe there is a better way to do this? 


Too lazy to look right now; consider

#include <string.h>
#include <stdlib.h>
#include <stdio.h>

#define NUM_OCTETTS 4

int iptoul (const char *ip, unsigned long *plong)
{
  char *next = NULL;
  const char *curr = ip;
  unsigned long tmp;
  int i, err = 0;

  *plong = 0;
  for (i = 0; i < NUM_OCTETTS; i++) {
    tmp = strtoul(curr, &next, 10);
    if (tmp >= 256
        || (tmp == 0 && next == curr))
    {
      err++;
      break;
    }

Forgot the obvious one:
     if (*next != '.' && i != (NUM_OCTETTS-1)) {
       err++;
       break;
     }

    *plong = (*plong << 8) + tmp;
    curr = next + 1;
  }

  if (err) {
    return 1;
  }
  else {
    return 0;
  }
}

int main (void)
{
    const char *ip = "124.15.22.102";
    unsigned long ret;

    if (0 == iptoul(ip, &ret))
      printf("%s -> %lu\n", ip, ret);

    return 0;
}

Cheers
 Michael



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