In Java I would do
System.out.println(new BigInteger(new byte[]{0,(byte)171,52,33}).intValue());
How would you do it in Perl?
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1It would be helpful if you showed the output of that.ysth– ysth2011-01-03 07:43:34 +00:00Commented Jan 3, 2011 at 7:43
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the output should be 11220001Eyal– Eyal2011-01-03 10:29:08 +00:00Commented Jan 3, 2011 at 10:29
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2 Answers
C:\Users\pgp\Documents\src\tmp>cat pack.pl
use Modern::Perl; # strict, warnings, v5.10 features
say unpack "N", pack "C4", 0, 171, 52, 33; # big endian
say unpack "V", pack "C4", 0, 171, 52, 33; # little endian
C:\Users\pgp\Documents\src\tmp>perl pack.pl
11220001
557099776
I can't remember what endianness Java specifies, but you can take your pick.
EDIT: as ysth helpfully points out, this has a 32-bit limit. I think there are pack options up to 64 bits, but no further. If you need arbitrary precision, his answer is better.
2 Comments
ysth
I was assuming the BigInteger part meant the need to support arbitrarily large numbers...
Philip Potter
@ysth yes, that's a reasonable interpretation, although the OP didn't specify explicitly.
I'm guessing you would want something like:
#!/usr/bin/perl
use strict;
use warnings;
use 5.010;
use Math::BigInt;
say Math::BigInt->new( '0x' . join('', map sprintf('%.2x', $_), 171, 52, 33) );
This converts the array elements to a hex string 0xab3421 and uses that to create a bigint.
