On Nov 24, Alexandre Checinski said:
I have lists that looks like this :
(A,B,C,1,5)
(A,B,C,1,2)
(A,B,2,1,2)
(A,B,C,D,1)
(A,B,Y)
And I would like to get something like this :
Here's how to do it without using eval().
use strict;
use warnings;
use Data::Dumper;
$Data::Dumper::Indent = 1;
my @lists = (
[qw( A B C 1 5 )],
[qw( A B C 1 2 )],
[qw( A B 2 1 2 )],
[qw( A B C D 1 )],
[qw( A B Y )],
);
my %hash;
for my $path (@lists) {
add_path([EMAIL PROTECTED], \%hash);
}
print Dumper(\%hash);
sub add_path {
my $path = shift;
if (@$path) { add_path($path, $_[0]{shift @$path}) }
else { $_[0] = 1 }
}
It works because of a nifty Perl behavior: the elements of @_ (the
arguments to a function) are *aliases* to the things passed to the
function. That means, by altering $_[0], you can alter the first argument
to a function.
In my code, I've shift()ed the @_ array, so altering $_[0] is actually
changing the *second* argument to the function. But anyway.
We send the add_path() function an array reference (the path) and a
reference to where to store the path. So for (A, B, C, 1, 5), we do:
add_path([A,B,C,1,5], \%hash);
The add_path() function checks to see if there are still elements in the
path arrayref. If there are, it removes the first part and uses it as a
key to the hash reference it was given and calls itself again:
add_path([B,C,1,5], $_[0]{A});
where $_[0] is \%hash, which calls
add_path([C,1,5], $_[0]{B});
where $_[0] is $hash{A}, which calls
add_path([1,5], $_[0]{C});
where $_[0] is now $hash{A}{B}, which calls
add_path([5], $_[0]{1});
where $_[0] is now $hash{A}{B}{C}, which calls
add_path([], $_[0]{5});
where $_[0] is now $hash{A}{B}{C}{1}...
Now, when the path is empty, instead of calling add_path() again, we set
$_[0] to 1.
This means that $hash{A}{B}{C}{1}{5} is set to 1.
This can be done WITHOUT a recursive function. Exercise to the reader.
--
Jeff "japhy" Pinyan % How can we ever be the sold short or
RPI Acacia Brother #734 % the cheated, we who for every service
http://www.perlmonks.org/ % have long ago been overpaid?
http://princeton.pm.org/ % -- Meister Eckhart
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