Is there a function to convert from binary to hexadecimal?

[SparceMatrix]

I've looked up all the permutations using "oct" and don't see a solution there. "hex" won't do it. If "sprintf" will do it, I can't tell from the documentation and "unpack" is even more obtuse. A staged conversion such as binary -> decimal -> hexadecimal won't do, it must be binary -> hexadecimal.

A simple question ... anyone?

 

[AMiSM]

You might use a hash containing all sixteen hex charactors (eg - $hash{1101}="B") and a regex to match the rightmost four charactors. Maybe something like:

$hex = "";
$_ = $bin;
$hex = $hash{/(0|1){4}$/} . $hex;
s/(0|1){4}$//;
(repeat as needed)

...not sure if precisely this will work, though. Haven't messed with hashes much. Anybody else want to flesh this out?

 

[cfmartin2000]

This'll work:

my %hex = (
0000 => '0',
0001 => '1',
0010 => '2',
0011 => '3',
0100 => '4',
0101 => '5',
0110 => '6',
0111 => '7',
1000 => '8',
1001 => '9',
1010 => 'A',
1011 => 'B',
1100 => 'C',
1101 => 'D',
1110 => 'E',
1111 => 'F',
);


my $x = '110111011111101111101000110000';


my $hexnumber = '';

while($x=~/(....)/g){

$hexnumber .= $hex{$1};

}

print $hexnumber . "\n";

 

[SparceMatrix]

Thanks all, but I bumped into a nice module over at ActiveState,

http://aspn.activestate.com/ASPN/CodeDoc/Math-BaseCalc/BaseCalc.html

I copy and pasted the example and with a few alterations, got exactly what I was looking for:

#!/usr/bin/perl

use Math::BaseCalc;

$calc1  = new Math::BaseCalc(digits=>[0,1]);
$calc63 = new Math::BaseCalc(digits=>[0..9,'a'..'z','A'..'Z']);

$MyHugeBinary =  "1" x 192;

$in_base_63 = $calc63->to_base( $calc1->from_base($MyHugeBinary) );

print $in_base_63;
print "   And the length is ... ". length( $in_base_63 ). "\n";
print "... as opposed to ". length($MyHugeBinary). "\n";

 

[AMiSM]

One thing I noticed is that your hex has to be a multiple of four, or you lose some digits. Maybe some leading zeros should be tacked on? How about changing the regex to

while($x=~/(.{1-4})/g){

...then the hex doesn't need to be a multiple.

Another problem is if regex's process from left to right, in which case, if the hex isn't a multiple of four, the clustering would be out of phase, and the result would be wrong.