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Re: Parsing a Tree to a Table.

by Discipulus (Curate)
on Dec 05, 2013 at 10:15 UTC ( #1065738=note: print w/ replies, xml ) Need Help??


in reply to Parsing a Tree to a Table.

hello,
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?
L*
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.


Comment on Re: Parsing a Tree to a Table.
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Re^2: Parsing a Tree to a Table.
by Athanasius (Monsignor) on Dec 05, 2013 at 11:17 UTC

    The behaviour here has nothing to do with when the increment operation is applied. $deep + 1 evaluates to 1 more than the current value of $deep, but leaves that value unchanged. ++$deep (or $deep++) changes the variable’s value by incrementing it, so that on the next iteration of the foreach loop, $deep is larger (by 1) than it was before. ++$deep is equivalent to $deep += 1 — the latter form makes the change to the variable explicit via the presence of an assignment operator.

    Hope that helps,

    Athanasius <°(((><contra mundum Iustus alius egestas vitae, eros Piratica,

      OMG what a stupid's moment passed to me!!
      thanks
      L*

      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.

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