Re: binary converter

[Limbic~Region]

perlnoobz,
I will re-write your code and explain it. I hope it makes sense and isn't more confusing to you.

#!/usr/bin/perl
use strict;
use warnings;

# Obtain a binary string from the user or die trying
print "Your binary input please => ";
chomp(my $bin = <STDIN>);
die "'$bin' is not binary\n" if $bin =~ /[^01]/;

# Binary numbers are easier to convert right to left
# We will reverse the numbers to do this

my @position = reverse split //, $bin;

# Now the right most value is in the first slot of the array
# And the left most value is in the last slot of the array

my $tot = 0;

# Normally when we loop over an array, we loop over its values
# In this case, looping over its indices allows us to do powers of 2

# $#position is the last index in the array

for my $idx (0 .. $#position) {
    $tot += $position[$idx] * 2 ** $idx;  # had a typo here with $digi
+t[$idx]
}

print "\n$tot\n";

[Jenda]

The position shoul be digit :-)

In either case you do not need the index nor the ** operator.

my @digit = reverse split //, $bin;

my $num = 0;
for my $bit (@digit) {
 $num = $num * 2 + $bit;
}
print "\n$num\n";

[Limbic~Region]

Jenda,
Or digit should have been position.

I wouldn't have written the code I did for myself - it was intended to be more instructional since the OP indicated not understanding the for loop. I guess I should have paid closer attention (updated now). Thanks!