so I know there are libraries that can do this for me but I want to learn pack / unpack.
my goal is I have a user input an ip address / subnet mask and then verify that it's valid.
one way i thought of doing it was "sprintf" and get a binary value of lets say 192.168.1.1 . that's an ok solution, but then i need to prepend the required amount of 0's t make it '8 bit'
that seems like a lot of unnecessary work when pack can put things in binary format. I used the N template i found http://perldoc.perl.org/functions/pack.html my first goal was to get an ip address, convert it to binary, then convert it back.
$ip = "192.168.1.1";
$bi = pack ("N*", $ip);
print unpack("N*",$bi),"\n";
and the output i got was 192 so obviously i don't understand what's going on here.
what exactly is going on here?
pack ("N*", $ip) takes an integer out of $ip and puts it into network byte order.
What you want is packing the 4 decimal octets of the IP address as binary. No need to fiddle with endianness as the IP address string is already in big endian (The highest order byte is already at the start of the string).
I also changed the * to a 4, IP addresses are always 4 octets long:
$ip = "192.168.1.1";
$bi = pack "C4", split('\.', $ip);
print join('.', unpack("C4",$bi)), "\n";
Related
why doesn't this bor bnot give the expected result in powershell?
To find the last address in an ipv6 subnet one needs to do a "binary or" and a "binary not" operation.
The article I'm reading (https://www.codeproject.com/Articles/660429/Subnetting-with-IPv6-Part-1-2) describes it like this:
(2001:db8:1234::) | ~(ffff:ffff:ffff::) = 2001:db8:1234:ffff:ffff:ffff:ffff:ffff
Where | is a "binary or" and
~ is a "binary not"
In powershell however, I try it like:
$mask = 0xffffffff
$someOctet = 0x0000
"{0:x4}" -f ($someOctet -bor -bnot ($mask) )
and I get 0000 instead of ffff
Why is this?
The tutorial is doing a -not of the entire subnet mask, so ff00 inverts to 00ff and similar for longer Fs and 0s; you aren't doing that, so you don't get the same results.
The fully expanded calculation that you show is doing this:
1. (2001:0db8:1234:0000:0000:0000:0000:0000) | ~(ffff:ffff:ffff:0000:0000:0000:0000:0000)
2. (2001:0db8:1234:0000:0000:0000:0000:0000) | (0000:0000:0000:ffff:ffff:ffff:ffff:ffff)
3. = 2001:db8:1234:ffff:ffff:ffff:ffff:ffff
Note how in step 1. to step 2, the not is inverting the pattern of Fs and 0s, switching the subnet mask around, and switching it around between the bit where the prefix ends and the bit where the host part begins.
Then step 3 or takes only the set bits from the left to keep those numbers the same (neither zero'd nor ffff'd), and all the set bits from the right (to ffff those, maxing them to the max IP address within that prefix).
In other words, it makes no sense to do this "an octet at a time". This is a whole IP address (or whole prefix) + whole subnet mask operation.
Where the tutorial says:
& (AND), | (OR), ~ (NOT or bit INVERTER): We will use these three bitwise operators in our calculations. I think everybody is familiar -at least from university digital logic courses- and knows how they operate. I will not explain the details here again. You can search for 'bitwise operators' for further information.
If you aren't very familiar with what they do, it would be worth studying that more, before trying to apply them to IP subnetting. Because you are basically asking why 0 or (not 1) is 0 and the answer is because that's how Boolean logic "or" and "not" work.
Edit for your comment
[math]::pow(2,128) is a lot bigger than [decimal]::maxvalue, so I don't think Decimal will do.
I don't know what a recommended way to do it is, but I imagine if you really wanted to do it all within PowerShell with -not you'd have to process it with [bigint] (e.g. [bigint]::Parse('20010db8123400000000000000000000', 'hex')).
But more likely, you'd do something more long-winded like:
# parse the address and mask into IP address objects
# which saves you having to expand the short version to
$ip = [ipaddress]::Parse('fe80::1')
$mask = [ipaddress]::Parse('ffff::')
# Convert them into byte arrays, then convert those into BitArrays
$ipBits = [System.Collections.BitArray]::new($ip.GetAddressBytes())
$maskBits = [System.Collections.BitArray]::new($mask.GetAddressBytes())
# ip OR (NOT mask) calculation using BitArray's own methods
$result = $ipBits.Or($maskBits.Not())
# long-winded way to get the resulting BitArray back to an IP
# via a byte array
$byteTemp = [byte[]]::new(16)
$result.CopyTo($byteTemp, 0)
$maxIP = [ipaddress]::new($byteTemp)
$maxIP.IPAddressToString
# fe80:ffff:ffff:ffff:ffff:ffff:ffff:ffff
Simple enough, I'd like to split a given IP address into netid (as defined by the netmask) and the hostid in Perl. Example:
$network = NetAddr::IP->new('192.168.255.255/29') || die "invalid space $_";
Now $network->mask returns 255.255.255.248. But there're no methods in NetAddr::IP to apply the mask to split the address into its netid and hostid portions in the /29 space.
NetAddr::IP::Util mentions the operators to do so, but it's documentation is a mess.
At least the netid can be extracted using Net::NetMask:
$netid = Net::Netmask->new('192.168.255.255/29')->base;
This yields 192.168.255.248. Again, no method to get the host portion 0.0.0.7. Maybe the best would be to pack/unpack the IPs into 32 bit int and then simply & them out. Then it would be easier to print the binary representations of IP addresses too, which I found can be really helpful for debugging and documentation purposes.
Use the hostmask() method
$host_wildcard = Net::Netmask->new('192.168.255.255/29')->hostmask;
I read some raw data from my device. This data contains the IP address as well but in a different format. As you know the IP address is generally written in the format a.b.c.d. However I have data of the format abcd given from the device. I need to get this in the format a.b.c.d How do I do this in a perl script?
Regards
First, let us split the hex string into substrings of two characters:
... split /..\K/, "c0a80001";
We treat each fragment as a hex string, and get the numeric value with the hex builtin:
... map hex, ...
Then, we join all numbers with a period:
join '.', ...
Combined:
my $ip = join '.', map hex, split /..\K/, "c0a80001";
print "$ip\n";
Output: 192.168.0.1. This is the usual text representation for an IPv4 address.
There are many ways. This inserts dots with substring.
map { substr($string,$_,0)='.' } (6,4,2);
Maybe you prefer regexes.
$string =~ s/[0-9a-f]{2}\K(?!\Z)/./g;
It really depends on the approach taken, but mostly you would need to escape with a backslash the dot from the IP address
a\.b\.c\.d
Some source code with be nice btw ...
I'm relatively inexperienced with Perl, but my question concerns the unpack function when getting the bits for a numeric value. For example:
my $bits = unpack("b*", 1);
print $bits;
This results in 10001100 being printed, which is 140 in decimal. In the reverse order it's 49 in decimal. Any other values I've tried seem to give the incorrect bits.
However, when I run $bits through pack, it produces 1 again. Is there something I'm missing here?
It seems that I jumped to conclusions when I thought my problem was solved. Maybe I should briefly explain what it is I'm trying do.
I need to convert an integer value that could be as big as 24 bits long (the point being that it could be bigger than one byte) into a bit string. This much can be accomplished using unpack and pack as suggested by #ikegami, but I also need to find a way to convert that bit string back into it's original integer (not a string representation of it).
As I mentioned, I'm relatively inexperienced with Perl, and I've been trying with no success.
I found what seems to be an optimal solution:
my $bits = sprintf("%032b", $num);
print "$bits\n";
my $orig = unpack("N", pack("B32", substr("0" x 32 . $bits, -32)));
print "$orig\n";
This might be obvious, but the other answers haven't pointed it out explicitly: The second argument in unpack("b*", 1) is being typecast to the string "1", which has an ASCII value of 31 in hex (with the most significant nibble first).
The corresponding binary would be 00110001, which is reversed to 10001100 in your output because you used "b*" instead of "B*". These correspond to the opposite "endian" forms of the binary representation. "Endian-ness" is just whether the most-significant bits go at the start or the end of the binary representation.
Yes, you're missing that different machines support different "endianness". And Perl is treating 1 like '1' so ( 0x31 ). So, you're seeing 1 -> 1000 (in ascending order) and 3 -> 1100.
"Wrong" depends on perspective and whether or not you gave Perl enough information to know what encoding and endianness you wanted.
From pack:
b A bit string (ascending bit order inside each byte, like vec()).
B A bit string (descending bit order inside each byte).
I think this is what you want:
unpack( 'B*', chr(1))
You're trying to convert an integer to binary and then back. While you can do that with pack and then unpack, the better way is to use sprintf or printf with the %b format:
my $int = 5;
my $bits = sprintf "%024b\n", $int;
print "$bits\n";
To go the other way (converting a string of 0s & 1s to an integer), the best way is to use the oct function with a 0b prefix:
my $orig = oct("0b$bits");
print "$orig\n";
As the others explained, unpack expects a string to unpack, so if you have an integer, you first have to pack it into a string. The %b format expects an integer to begin with.
If you need to do a lot of this on bytes, and speed is crucial, you could build a lookup table:
my #binary = map { sprintf '%08b', $_ } 0 .. 255;
print $binary[$int]; # Assuming $int is between 0 and 255
The ord(1) is 49. You must want something like sprintf("%064b", 1), although that does seem like overkill.
You didn't specify what you expect. I'm guessing you're expecting 00000001.
That's the correct bits for the byte you provided, at least on non-EBCDIC systems. Remember, the input of unpack is a string (mostly strings of bytes). Perhaps you wanted
unpack('b*', pack('C', 1))
Update: As others have pointed out, the above gives 10000000. For 00000001, you'd use
unpack('B*', pack('C', 1)) # 00000001
You want "B" instead of "b".
$ perl -E'say unpack "b*", "1"'
10001100
$ perl -E'say unpack "B*", "1"'
00110001
pack
I have a variable that contains a 4 byte, network-order IPv4 address (this was created using pack and the integer representation). I have another variable, also a 4 byte network-order, subnet. I'm trying to add them together and add one to get the first IP in the subnet.
To get the ASCII representation, I can do inet_ntoa($ip&$netmask) to get the base address, but it's an error to do inet_ntoa((($ip&$netmask)+1); I get a message like:
Argument "\n\r&\0" isn't numeric in addition (+) at test.pm line 95.
So what's happening, the best as I can tell, is it's looking at the 4 bytes, and seeing that the 4 bytes don't represent a numeric string, and then refusing to add 1.
Another way of putting it: What I want it to do is add 1 to the least significant byte, which I know is the 4th byte? That is, I want to take the string \n\r&\0 and end up with the string \n\r&\1. What's the simplest way of doing that?
Is there a way to do this without having to unpack and re-pack the variable?
What's happening is that you make a byte string with $ip&$netmask, and then try to treat it as a number. This is not going to work, as such. What you have to feed to inet_ntoa is.
pack("N", unpack("N", $ip&$netmask) + 1)
I don't think there is a simpler way to do it.
Confusing integers and strings. Perhaps the following code will help:
use Socket;
$ip = pack("C4", 192,168,250,66); # why not inet_aton("192.168.250.66")
$netmask = pack("C4", 255,255,255,0);
$ipi = unpack("N", $ip);
$netmaski = unpack("N", $netmask);
$ip1 = pack("N", ($ipi&$netmaski)+1);
print inet_ntoa($ip1), "\n";
Which outputs:
192.168.250.1