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Perl: the Markov chain saw

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about the rosetta code: it says nor the tree or the output are important. For me a tree is always an HoH and this remember me an ancient piece of code i wrote long time ago (little refurbished to run under strict... now i'm little wiser ;=) ).
use strict; use warnings; my %h1=('qwqw' => {'qw' => 'qww','df' => 'dfff','C1' => {'A2' => 'a2', +'B2' => 'b2','C2' => 'c2END'}},); my %h2=( 'AAA' => 'aaa', 'BBB' => 'bbb', 'CCC' => {'A1' => 'a1','B1' => 'b1','C1' => {'A2' => 'a2','B2' => 'b2' +,'C2' => 'c2END'}}, 'DDD' => 'ddd', 'EEE' => \%h1, 'FFF' => 'sdsfsds', ); &ddump(\%h2); sub ddump { my $ref = shift; my $deep = shift||0; foreach my $k (sort keys %{$ref}) { if (ref( ${$ref}{$k})) {print "\t" x $deep."$k =>\n"; &dd +ump (${$ref}{$k}, ($deep+1))} else {print "\t" x ($deep)."$k => ${$ref}{$k}\n";} } }

PS I noticed something I do not understand: if you change $deep+1 whit ++$deep the beahaviur change: why? in docs i learn:
Note that just as in C, Perl doesn't define when the variable is incremented or decremented. You just know it will be      done sometime before or after the value is returned. This also means that modifying a variable twice in the same statement will lead to undefined behavior.
Is this the case?
There are no rules, there are no thumbs..
Reinvent the wheel, then learn The Wheel; may be one day you reinvent one of THE WHEELS.

In reply to Re: Parsing a Tree to a Table. by Discipulus
in thread Parsing a Tree to a Table. by choroba

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