Hi,
I was reading Higher Order Perl[1] last night and, dishearteningly,
got stuck on chapter one.
Mark Dominus offers the following as a binary string conversion
example of how recursion can work.
use strict;
use warnings;
my $bstr = binary(37);
print "$bstr\n"; # prints 100101
sub binary {
my ($n) = @_;
return $n if $n == 0 || $n == 1;
my $k = int($n/2);
my $b = $n % 2;
my $E = binary($k);
return $E . $b;
}
The algorithm works perfectly but my understanding of it's workings is amiss.
When I look at this I see $E initialised and then concatenate with the
the modulus of
37 % 2 =1
18 % 2 = 0
9 % 2 = 1
4 % 2 = 0
2 % 2 = 0
That by my reckoning is 10100. Almost the reverse of the answer but I
am obviously wrong and I can't see how the final expressions:
my $E = binary($k);
return $E . $b;
work to give the answer.
Can someone more enlightened than me give me some guidence.
Thanx,
Dp.
PS: Has anybody heard from Rob Dixon, he hasn't been on the list for ages.
1. http://hop.perl.plover.com/book
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