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Question : inet_pton

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  Question :  inet_pton joao <joao82@simpmail.com.br> - 2011-12-07 21:54 +0000
    Re: Question :  inet_pton Ian Collins <ian-news@hotmail.com> - 2011-12-08 11:02 +1300
    Re: Question :  inet_pton Ike Naar <ike@sverige.freeshell.org> - 2011-12-07 22:45 +0000
      Re: Question :  inet_pton Ike Naar <ike@sverige.freeshell.org> - 2011-12-07 22:48 +0000

#14377 — Question : inet_pton

Dears friends

Do you give me an advise for expand IPv6 adresses with inet_pton() ? I 
need expand then increment in 1 and display.

I am having this :

unsigned char addr[16] = inet_pton(AF_INET6,"1234:12::ffff",&addr) ;

but he no work.

Many Thanks !

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#14378

On 12/ 8/11 10:54 AM, joao wrote:
> Dears friends
>
> Do you give me an advise for expand IPv6 adresses with inet_pton() ? I
> need expand then increment in 1 and display.

The best advice is to ask on comp.unix.programmer.  inet_pton() is 
defined in the Unix specification.

-- 
Ian Collins

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#14380

On 2011-12-07, joao <joao82@simpmail.com.br> wrote:
> Do you give me an advise for expand IPv6 adresses with inet_pton() ? I 
> need expand then increment in 1 and display.
>
> I am having this :
>
> unsigned char addr[16] = inet_pton(AF_INET6,"1234:12::ffff",&addr) ;
>
> but he no work.

Did you read the documentation for inet_pton?
The return value of inet_pton is not an array, but an integer
that specifies whether the second argument was a valid address (1)
or not (0 or -1). Also, the & in &addr is redundant because addr is
an array.  You probably want something like:

#include <stdio.h>
#include <arpa/inet.h>

int main(void)
{  
  unsigned char addr[16];
  if (1 == inet_pton(AF_INET6, "1234:12::ffff", addr))
  {
    /* valid address, print the converted value */
    size_t i;
    for (i = 0; i != sizeof addr; ++i)
    {
      printf("%0x", addr[i]);
    }
    printf("\n");
  }
  return 0;
}

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#14381

On 2011-12-07, Ike Naar <ike@sverige.freeshell.org> wrote:
>       printf("%0x", addr[i]);

the format should be "%02x" (with "%0x" the output will be
a bit confusing).

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