Re: Bitwise operators

Ok here's what I've done, which works -

#!/usr/bin/perl
#converter.plx
use warnings;
use strict;
print "Please input a decimal no. to convert to binary : ";
chomp (my $value = <STDIN>);
print "$value & 1",  "$Value & 2",  "$value & 4";
print "$value & 8",  "$value & 16", "$value & 32";
print "$value & 64", "$value & 128\n";
[download]

but I've started the next chapter which is on arrays and hashes, so although I'm not sure how to do it yet, I'm thinking there's a probably a shorter way? Paul.

Comment on Re: Bitwise operators Download Code

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Re^2: Bitwise operators by liverpole (Monsignor) on Dec 25, 2006 at 19:34 UTC
Hi again, thevoid, I just now read this earlier post of yours. print "$value & 1", "$Value & 2", "$value & 4"; The problem you're having is that you can only interpolate a certain amount within strings, and doing bit-wise arithmetic is a little too much. You could do this if you used printf like so: `printf "%d, %d, %d, ", $value & 1, $value & 2, $value & 4;` [download] And you could even convert to one or zero values by applying the appropriate shifts: `printf "%d, %d, %d, ", ($value & 1), ($value & 2) >> 1, ($value & 4) > +> 2;` [download] s''(q.S:$/9=(T1';s;(..)(..);$..=substr+crypt($1,$2),2,3;eg;print$..$/	[reply] [d/l] [select]
Re^3: Bitwise operators by thevoid (Scribe) on Dec 25, 2006 at 22:14 UTC
Hi, just spotted this! That's pretty nifty also : )	[reply]