sub bintodec {
        unpack("N", pack("B32", substr("0" x 32 . shift, -32)));
    }
[download]

And this seems to work for a dectobin, although there's probably a faster way w/o using join:

    sub dectobin {
        join '', unpack "B32", pack "N", shift;
    }
[download]

   print bin2dec(0b1101) . "\n";
   print dec2bin(13) . "\n";

   sub bin2dec {
       return(sprintf "%d",shift);
   }

   sub dec2bin {
       return(sprintf "%b",shift);
   }
[download]

   perl -e 'print oct("0b1101") . "\n";'
[download]

Ouch. Your bin2dec is doing absolutely nothing but making sure it's an integer.

You really need to think in terms of "numbers" being processed, not "decimal numbers". The value 47 internally is not in decimal or binary, conceptually. It's just forty seven things. So $a = 47 starts with a decimal 47 as I type it in, then converts that to the concept of 47, and assigns that to $a.

So what we really have are building blocks:

$decimal = sprintf "%d", $num; # or trivially removed, since automatic
$octal = sprintf "%o", $num;
$hex = sprintf "%h", $num;
$binary = sprintf "%b", $num; # 5.6 only
$num = 0+ $decimal; # or trivially removed, since automatic
$num = oct($octal);
$num = unpack "N", pack "H*", $hex;
$num = unpack "N", pack "B*", $binary;
[download]

Then combine as needed.

-- Randal L. Schwartz, Perl hacker

Ouch. Your bin2dec is doing absolutely nothing but making sure it's an integer.

Ehh.. but it does the job in question, right? Converting a supplied binary number to decimal.

update: OK, maybe that's not the definition of "decimal". Everyone happy if we rename the sub to bin2int? :)