Counting keys in a referenced hash

logie17 has asked for the wisdom of the Perl Monks concerning the following question:

Replies are listed 'Best First'.
Re: Counting keys in a referenced hash by Sidhekin (Priest) on Dec 02, 2006 at 20:06 UTC
You are using symbolic references. Your second hash is in fact %3. If you had been using strict, you'd have known: `Can't use string ("3") as a HASH ref while "strict refs" in use ...` [download] You would get a real hashref (which you are obviously trying to get), if you had used the hash constructor, `{...}`, instead of the parentheses. But the real lesson you should learn here, that might save you both time and future embarrassment, is the same old: `use strict; use warnings;` [download] `print "Just another Perl ${\(trickster and hacker)},"` The Sidhekin proves Sidhe did it!	[reply] [d/l] [select]
Re: Counting keys in a referenced hash by tomazos (Deacon) on Dec 02, 2006 at 20:07 UTC
The second one should be written with curly brackets not parenthesis. `$test = { test1 => '1', test2 => '2', test3 => '3'};` [download] Otherwise you are trying to assign a list to a scalar. `use warnings` would have told you this: `Useless use of constants in void context at test.pl line 12.` [download] This is referring to the members of the list that were discarded when it was assigned to a scalar. Everything but the last 3 was discarded. Further `use strict` would have told you: `Can't use string ("3") as a HASH ref while "strict refs" in use at tes +t.pl line 19.` [download] $test holds the number 3. When you try to dereference the number 3 (a la `%{$test}` it converts it to a string "3" and then looks up the hash called "%3" in the symbol table. strict doesn't allow such shenanigans. The curly brackets in the correct example indicate an anonymous hash reference. References are scalars. So it assigns to $test just fine. Don't study too hard though - the confusion between references-to-objects and the objects themselves will be going away in Perl 6. (or getting worse, depending how you look at it, :) ) -Andrew.	[reply] [d/l] [select]
Re: Counting keys in a referenced hash by mickeyn (Priest) on Dec 02, 2006 at 20:08 UTC
$test doesn't refer to a hash ... it actually gets the last element from the list which is '3'. apply the following change: `$test = { test1 => '1', test2 => '2', test3 => '3'};` [download] HTH, Mickey	[reply] [d/l]
Re: Counting keys in a referenced hash by ocs (Monk) on Dec 02, 2006 at 20:11 UTC
Hi Logan, your second thing what you consider to be a hashref isn't really one. You have to create a anonymous hashref with these curly brackets "{". This is what you mean: `$test = { test1 => '1', test2 => '2', test3 => '3'};` [download]	[reply] [d/l]
Re^2: Counting keys in a referenced hash by logie17 (Friar) on Dec 02, 2006 at 20:16 UTC
Thank you everyone. All I have to say is "Doh!". Thanks again Logan	[reply]
Re^3: Counting keys in a referenced hash by aufflick (Deacon) on Dec 04, 2006 at 05:10 UTC
Actually there is something else you could say, and that's that you will heed the advice to always 'use strict' and 'use warnings'.	[reply]
Re: Counting keys in a referenced hash by dimwit (Acolyte) on Dec 03, 2006 at 21:33 UTC
Not to nitpick but is this homework or just persoanl curiosity? If it is homework I think you should so state. Personally I am willing to help in either case but I do like to know. Other than that I can only say what everyone else has said: `use warnings; use strict;` [download] There is no better habit to develope when writing Perl. xenchu the dim (ie. dimwit) Well, 'splain it to me!	[reply] [d/l]
Re^2: Counting keys in a referenced hash by xiaoyafeng (Deacon) on Dec 04, 2006 at 07:08 UTC
I understood your opnion,but a newbie might be confused.:-)	[reply]
Re^2: Counting keys in a referenced hash by logie17 (Friar) on Dec 06, 2006 at 02:48 UTC
Xenchu, This is not for a homework assignment other than I'm a perl hobbyist and learned a very embarrasing lesson today in the monastary. Thanks though for the feedback, Logan s;;5776?12321=10609$d=9409:12100$xx;;s;(\d*);push @_,$1;eg;map{print chr(sqrt($_))."\n"} @_;	[reply]
Re: Counting keys in a referenced hash by Anonymous Monk on Dec 03, 2006 at 13:50 UTC
%test is a hash; $test is a scalar with value 3 Possibly, `$test = { test1 => '1', test2 => '2', ... };` was meant. Follow-up question: Why does $test have a value of 3 and not 6? ... since `( test1 => '1', test2 => '2', ... )` is the same as `( 'test1, '1', 'test2', '2', ... )`	[reply] [d/l] [select]
Re^2: Counting keys in a referenced hash by Locutus (Beadle) on Dec 03, 2006 at 16:11 UTC
Dear Anonymus Monk, `$test` doesn't have a value of 6 because unlike an array variable a list literal doesn't evaluate to the number of elements if used in scalar context but to its last element. Try `#!/usr/bin/perl use strict; use warnings; my $scalar = qw/one two three/; my @array = qw/one two three/; print '$scalar is ', $scalar, "\n"; # prints + 'three' print '@array in scalar context is ', scalar @array, "\n"; # prints + 3` [download] for example. Perl supposes you know the length of a literally coded (i.e. hard-coded) list anyway and tries to offer you something more useful when it discovers a "Useless use of a constant in void context"...	[reply] [d/l] [select]
Re: Counting keys in a referenced hash by jonathanu (Novice) on Dec 05, 2006 at 21:09 UTC
I think that your second assignment is not doing what you expect: `$test = ( test1 => '1', test2 => '2', test3 => '3');` [download] This assigns the last value in the list to $test. Later, when you dereference it, treating it like a reference to a hash, you end up with an empty hash. Probably you meant to do this instead: `$test = { test1 => '1', test2 => '2', test3 => '3'};` [download] The curly braces give you a reference to a hash. This should give you the expected result. -JonathanU	[reply] [d/l] [select]


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