hash with a hash

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

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Re: hash with a hash by almut (Canon) on Jun 04, 2007 at 09:02 UTC
You probably want `$tests{2} = \%blah2;` [download] (i.e., you need to assign a hashref when you're in scalar context... Otherwise, you get the number of used/allocated buckets of the hash — which is why you see the error 'Can't use string ("2/8") as a HASH ref while "strict refs" in use...')	[reply] [d/l]
Re^2: hash with a hash by perlfan (Parson) on Jun 04, 2007 at 19:34 UTC
Yeah, `$tests{2} = \%blah2;` [download] is the same as `$tests{2} = { condition => "AND", terms => [ "foo", "bar" ], };` [download] One thing that you should note is that when explicitly defining hashes and arrays, one can use "`()`"'s. `my @array = ( 1, 2, 3 ); my %hash = ( A => 1, B => 2, C => 3, );` [download] However, anonymous arrays are defined with "`[]`"'s `my $array_ref = [ 1, 2, 3 ]; my @array = @{$array_ref}; # becomes an explicit array` [download] and anonymous hashes are defined with "`{}`"'s. `my $hash_ref = { A => 1, B => 2, C => 3, }; my %hash = %{$hash_ref}; # becomes an explicit hash` [download]	[reply] [d/l] [select]
Re: hash with a hash by andreas1234567 (Vicar) on Jun 04, 2007 at 09:32 UTC
Use Data::Dumper to inspect your datastructure. It probably is not what you think: `$ perl use strict; use warnings; my %blah1 = (); $blah1{condition} = ""; $blah1{terms} = [ "foobar" ]; my %blah2 = (); $blah2{condition} = "AND"; $blah2{terms} = [ "foo", "OR bar" ]; my %tests = (); $tests{1} = %blah1; $tests{2} = %blah2; use Data::Dumper; print Dumper(\%tests); __END__ $VAR1 = { '1' => '2/8', '2' => '2/8' };` [download] ++ for using strict and warnings. Read Perl Data Structures Cookbook for more on hashes of hashes. -- print map{chr}unpack(q{A3}x24,q{074117115116032097110111116104101114032080101114108032104097099107101114})	[reply] [d/l]
Re: hash with a hash by nferraz (Monk) on Jun 04, 2007 at 10:02 UTC
You can't store a hash inside of a hash, or an array inside of an array. But you can store references. Please read the perl data structures cookbook.	[reply]
Re: hash with a hash by 13warrior (Acolyte) on Jun 04, 2007 at 10:53 UTC
If you want to use a hash with another hash then you got to use references. $tests{2} = %blah2; You should use ref like $test{2} = \%blah2; Please go through references do use complex data structures.	[reply]
Re: hash with a hash by Anonymous Monk on Jun 04, 2007 at 18:43 UTC
Hashes can only store scalar variables. If you want to have a hash of hashes or hash of lists of hashes, etc. you need to use references. You can do something like this: my %foo = ( one => 'apple', two => 'orange' ); my %bar = ( foo => \%foo ); when you do \%foo you take a reference to %foo hash, and the reference is a scalar, which can be stored in a hash. in your case lines: $tests{1} = %blah1; $tests{2} = %blah2; do not do what you want, you need to re-write them like so: $tests{1} = \%blah1; $tests{2} = \%blah2; Here's a little test prog, try it: #!/usr/bin/perl -w use strict; use Data::Dumper; my %foo = ( one => 'apple', two => 'orange' ); my %bar = ( foo => \%foo ); # wrong $bar{1} = %foo; print Dumper(\%bar);	[reply]
Re: hash with a hash by TGI (Parson) on Jun 04, 2007 at 23:18 UTC
I've said it before, and I'll say it again, don't use the ampersand (&) sigil on your subroutine calls. It can have some nasty unintended consequences. See perlsub for more details. TGI says moo	[reply]

print map{chr}unpack(q{A3}x24,q{074117115116032097110111116104101114032080101114108032104097099107101114})