forming an ipv6 address string from unsigned char array

[sam.barker0@gmail.com]

Hi,
How can I convert an unsigned char array containing IPV6 address into
a string
Eg
if arrray contains
20 01 05 03 a8 3e 00 00 00 00 00 00 00 02 00 30

Then the address is
Addr: 2001:0503:a83e: 0000:0000:0000: 0002:0030

I tried to do write like below but then its completely wrong.Is there
any cast for hex
for(int i=0;i<=15;i++)
{
oss << static_cast<uns igned int>(*(Rec.GetS tart()+i));

Cheers,
Sam

[Oncaphillis]

Re: forming an ipv6 address string from unsigned char array

sam.barker0@gma il.com wrote:

Hi,
How can I convert an unsigned char array containing IPV6 address into
a string
Eg
if arrray contains
20 01 05 03 a8 3e 00 00 00 00 00 00 00 02 00 30
>
Then the address is
Addr: 2001:0503:a83e: 0000:0000:0000: 0002:0030
>
I tried to do write like below but then its completely wrong.Is there
any cast for hex
for(int i=0;i<=15;i++)
{
oss << static_cast<uns igned int>(*(Rec.GetS tart()+i));
>
Cheers,
Sam

How's about this ?

<snip>

#include <iostream>
#include <iomanip>
#include <sstream>

const unsigned char * cs =
(const unsigned char *)
"20 01 05 03 a8 3e 00 00 00 00 00 00 00 02 00 30";

int main() {
std::stringstre am is((const char*)cs);
std::stringstre am os;

int i;

for(i=0;i<16 && is;i++) {
unsigned int n;

is >std::hex >n;

if(!is || n 0xff) {
std::cerr << "failed to read (valid) number";
return -1;
}

os << (i % 2 || i==0 ? "" : ":")
<< std::hex << std::setw(2) << std::setfill('0 ') << n;
}

if(i!=16) {
std::cerr << "failed to eat up string" << std::endl;
return -1;
}

std::cerr << os.str() << std::endl;
}

</snip>

Hope that helps

o.

[sam.barker0@gmail.com]

Re: forming an ipv6 address string from unsigned char array

Hi,
I am sorry.I made a mistake.

arrray contains
32 01 05 03 168 62 00 00 00 00 00 00 00 02 00 48

Then the address is
Addr: 2001:0503:a83e: 0000:0000:0000: 0002:0030

The array is unsigned char type

[sam.barker0@gmail.com]

Re: forming an ipv6 address string from unsigned char array


Hi,
I have come up with the solution like this.
sprintf(tempstr ing, "%x:%x:%x:%x:%x :%x:%x:%x",hton s(*((unsigned short
*)(Rec))),htons (*((unsigned short *)(Rec+2))),hto ns(*((unsigned short
*)(Rec+4))),hto ns(*((unsigned short *)(Rec+6))),hto ns(*((unsigned
short *)(Rec+8))),hto ns(*((unsigned short *)(Rec
+10))),htons(*( (unsigned short *)(Rec+12))),ht ons(*((unsigned short *)
(Rec+14))));

Is there a better looking solution.
CHeers

[Oncaphillis]

Re: forming an ipv6 address string from unsigned char array

sam.barker0@gma il.com wrote:

Hi,
I am sorry.I made a mistake.
>
arrray contains
32 01 05 03 168 62 00 00 00 00 00 00 00 02 00 48
>
Then the address is
Addr: 2001:0503:a83e: 0000:0000:0000: 0002:0030
>
The array is unsigned char type

Hmm.. your first request wasn't very detailed.
So it's an array of unsigned char (which serve
as a octet here) which hold the ipv6 address.
NOT an array of char which hold the address
(encoded in *ASCII*).

So it seems like your working on the in6_addr
struct.

And you're mixing up decimal and hex representation
here. So 32 is supposed to mean 0x20 and
168 == 0xa8 and 62 == 0x3e etc.

O.

[Oncaphillis]

Re: forming an ipv6 address string from unsigned char array

sam.barker0@gma il.com wrote:

Hi,
I have come up with the solution like this.
sprintf(tempstr ing, "%x:%x:%x:%x:%x :%x:%x:%x",hton s(*((unsigned short
*)(Rec))),htons (*((unsigned short *)(Rec+2))),hto ns(*((unsigned short
*)(Rec+4))),hto ns(*((unsigned short *)(Rec+6))),hto ns(*((unsigned
short *)(Rec+8))),hto ns(*((unsigned short *)(Rec
+10))),htons(*( (unsigned short *)(Rec+12))),ht ons(*((unsigned short *)
(Rec+14))));
>
Is there a better looking solution.
CHeers

have a look at

http://www.opengroup.org/onlinepubs/000095399/functions/inet_ntop.html

I guess it solves your problem in a portable way.

Although it doesn't have to do with C++ ;-)

O.