perlnoobz has asked for the wisdom of the Perl Monks concerning the following question:

i do not understand the foreach loop..can anyone explain to me what is it doing? 

 #!/usr/bin/perl -w
print "Your binary input please =>";
$bin = <STDIN>; #read in user input 
chomp $bin;

# use of array to store $bin and split it
@digits = reverse(split('',$bin)); #reverse and split $bin, then assig
+n to an array
$result = 0;

foreach $pos (0..$#digits) {
$result += $digits[$pos] * 2 ** $pos; 
} 

print $result;

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Re: binary converter
by Limbic~Region (Chancellor) on Apr 22, 2013 at 15:30 UTC
    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";

    [download]

    Cheers - L~R

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      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";

      [download]

      Jenda
      Enoch was right!
      Enjoy the last years of Rome.

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[d/l]
[select]
        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!

        Cheers - L~R

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Re: binary converter
by toolic (Bishop) on Apr 22, 2013 at 15:36 UTC
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Re: binary converter
by CountOrlok (Friar) on Apr 22, 2013 at 15:18 UTC
    Try the code with a simple binary number, e.g. 100
    @digits will be (0,0,1). And $result will be 0*2^0 + 0*2^1 + 1*2^2 = 4.
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