Convert an ip address to long value

[kyle.tk]

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?

-kyle

[Michael Mair]

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;
}
*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
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Michael Mair]

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

--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Eric Sosman]

kyle.tk wrote On 01/31/06 15:27,:

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?

You could use sscanf(). Unfortunately, it's hard
to persuade sscanf() to reject things like "123.0.0.-1"
or " 127. 0. 0. 1".

Another possibility would be to use isdigit() to
check the first character of each field and then strtoul()
to convert it, then check that strtoul() stopped on a '.'
(or on a '\0' the final time). You might also check that
the stopping point wasn't too far from the start, so as
to reject "00000000000012 7.0.0.000000000 000000001".

--
Er*********@sun .com

[Old Wolf]

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 ";
That \0 is not necessary.
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);
Well, let's work through this loop.
Start: i = 11, j = 10.
Iter 1: j => 9. ip[9] == '.' so i => 9.
Iter 2: j => 8. i => 8. tmp[8] = ip[8] = '1'.
Iter 3: j => 7. ip[7] == '.' so i => 6.
Iter 4: j => 6. i => 5. tmp[5] = ip[6] = '8'.
Iter 5: j => 5. i => 4. tmp[4] = ip[5] = '6'.
Iter 6: j => 4. i => 3. tmp[3] = ip[4] = '1'.
Iter 7: j => 3. ip[3] = '.', so i => 3.
Iter 8: j => 2. i => 2. tmp[2] = ip[2] = '2'.
Iter 9: j => 1. i => 1. tmp[1] = ip[1] = '9'.
Iter10: j => 0. i => 0. tmp[0] = ip[0] = '1'.
Iter11: j => -1. Undefined behaviour, reading ip[-1].
Maybe will try and write to tmp[-1].

So you end up with tmp = "1921680010 00" and maybe having
written a buffer overflow before the start of tmp.

Hopefully it's now clear that your loop structure is pretty
weak and error-prone. A much better approach here is
to read forwards through ip, computing as you go, you
don't need to create a new array.
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];

tmp[0] is '1'. '1' * 100 will probably give you a value of 4900.
You mean to write: (tmp[0] - '0') * 100 , etc.

[kyle.tk]

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;
}
*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
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

Alright, after looking at that post I came up with this. I didn't use
yours verbatim because I have not yet implemented the stroul function.

Heres what I have got now. Critique?

unsigned long iptol(const char *ip){
unsigned long value = 0; /* Total Value */
unsigned char ocet = 0; /* Ocet Value */
int i = strlen(ip) - 1; /* Index in ip */
int m = 1; /* Ocet multiplier */
int j;
for (j=0; j < 4; j++){
while ( ip[i] != '.' && ip[i] != '\0' ){
ocet += m * (ip[i] - 48);
m *= 10;
i--;
}
value += (ocet << (8 * j));
ocet = 0;
m = 1;
i--;
};
return value;
}
-kyle

[Rod Pemberton]

"kyle.tk" <ky*****@gmail. com> wrote in message
news:11******** **************@ f14g2000cwb.goo glegroups.com.. .

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?

Yes, this is an alternate method, using more string processing. It compiles
and works for DJGPP and OW1.3:
#include <stdio.h>
#include <string.h>

unsigned long iptoul(char *ip)
{
char i,*tmp;
unsigned long val=0, cvt;

tmp=strtok(ip," .");
for (i=0;i<4;i++)
{
sscanf(tmp,"%lu ",&cvt);
val<<=8;
val|=(unsigned char)cvt;
tmp=strtok(NULL ,".");
}
return(val);
}

int main(void)
{

char *ip="124.15.22. 102";

unsigned long ret;

printf("%s\n",i p);
ret=iptoul(ip);
printf("%08lx\n ",ret);
return(0);
}
Rod Pemberton

[websnarf]

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;
}
I don't see any any conversion from character to digits (via a "- '0'"
kind of operation.) So I can't see how this is possibly supposed to
work. Furthermore, it is possible and legal to specify IP addresses
in which each portion might exceed 255, and therefore might have more
than 3 digits -- its supposed to just wrap around.
int main(void){
char *ip = "124.15.22.102" ;
iptol(ip);
}

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

This is off the top of my head:

unsigned long iptol (const char *ip) {
unsigned long ipl, b;
int i, m;
for (b = ipl = 0ul, m=4, i = 0; ; i++) {
switch (ip[i]) {
case '0': case '1': case '2': case '3': case '4':
case '5': case '6': case '7': case '8': case '9':
b = b*10 + ip[i] - '0';
break;
default:
ipl = (ipl << 8ul) + b;
if ('.' != ip[i] || 0==(--m)) return ipl;
b = 0;
break;
}
}
}

--
Paul Hsieh
http://www.pobox.com/~qed/
http://bstring.sf.net/

[CBFalconer]

"kyle.tk" wrote:

.... snip ...
Alright, after looking at that post I came up with this. I didn't use
yours verbatim because I have not yet implemented the stroul function.

strtoul() is a standard function. You don't need to implement it.

--
"The power of the Executive to cast a man into prison without
formulating any charge known to the law, and particularly to
deny him the judgement of his peers, is in the highest degree
odious and is the foundation of all totalitarian government
whether Nazi or Communist." -- W. Churchill, Nov 21, 1943

[kyle.tk]

CBFalconer wrote:

strtoul() is a standard function. You don't need to implement it.

I know it is a standard function in libc. The platform I am targeting
does not yet have a fully ported libc.

-kyle