Converting binary to decimal in PERL

Say
$x_bin="0001001100101"
$x_dec=613
so you need to convert $x_bin into $x_dec. Use oct():
$x_dec = oct("0b".$x_bin);

Quoting from man perlfunc:oct EXPR oct Interprets EXPR as an octal string and returns the corresponding value. (If EXPR happens to start off with "0x", interprets it as a hex string. If EXPR starts off with "0b", it is interpreted as a binary string. Leading whitespace is ignored in all three cases.)

You can also use Bit::Vector:
use Bit::Vector;
my $v = Bit::Vector->new_Bin( 32, '0001001100101' );
print "dec: ", $v->to_Dec(), "\n";

I have to get input for the user and spit out the resulting decimal number.  I'm not sure how to implement your description into my code.  It's kind of confusing.

What are you confused about?

If you post your code, we can more easily determine how to integrate the above code into it.

Here is what Ive come up with so far.  When I am entering 20 digit binary numbers tho the hex numbers are a little off.  For example.  Enter a binary number up to 20 digits: 11111111111111111111
11111111111111111111 is 1048577 in decimal
The correct answer should be 1048575.

#!/usr/bin/perl 

use warnings;
use strict;
print "\n";
print 'Enter a binary number up to 20 digits: ' ;
chomp ( my $num = <> );

my $pow=1;
my $dec=0;
{
my $bin=$num;
until($num==0)
{
my $bit=$num%10;
$dec=$dec+($bit*$pow);
$pow=$pow*2;
$num=$num/10;
}
print("$bin is $dec in decimal\n");
$pow=1;
$dec=0;
}

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To see the problem with your method, add
printf "%.9f\n",$num;
to your loop

hmmm... that was my solution but no points for me...  :-)

OZO

I am looking at my code for this problem again and need some help.  Looking at my original code, not your code using oct, what do I need to correct to get the correct results?  I get the right answer when using eleven 1's, but not when using twenty 1's.

You would need to treat $num as a string rather than as a decimal number.
As a decimal number, twenty 1's exceeds the precision of the floating point numbers on your machine.

Either that or use an extended precision module.

It's been awhile since I have worked in perl.  How do I modify this to make it a string?

#!/usr/bin/perl

use warnings;
use strict;
print "\n";
print 'Enter a binary number up to 20 digits: ' ;
chomp ( my $num = <> );
{
my $pow=1;
my $dec=0;
my $bin=$num;
until(!$num)
{
my $bit=chop $num;
$dec=$dec+($bit*$pow);
$pow*=2;
}
print("$bin is $dec in decimal\n");
}