You're right - it can't be done without saving the name of the name of the variable and some additional jiggery pokery. At least it can't be done with a hard reference like this. The reason it can't be done is because a hard reference is just and address - so the "thingie" that the address points to might have no varable name, such as an anonymous hash:

$hashref = {'Anonymous' => 'Hash'}; #hash has no varable
                                    #name - just reference

[download]

Additionally, a varable might have more than one name, if it has aliases.

Back in Perl 4 (which I've never used) there were no hard refs, so you had to use variable - name - plus eval tricks to do reference type stuff. This might do what you want it to (warning - untested code):

#usr/bin/perl  -w
use strict;
my %hash1 = ('A' => 1);
my $hashname1 = 'hash1';

my %hash2 = ('B' => 2);
my $hashname2 = 'hash2';

eval qq (print qq(Hashname: $$hashname1 Values:); print join ",",sort 
+keys %{$hasname1})" );
eval qq (print qq(Hashname: $$hashname1 Values:); print join ",",sort 
+keys %{$hasname1}) );

[download]

But these kinds of eval tricks I'm not used to - and you probably shouldn't use them anyway. Also, you can't print out a memory address here, unless you add in a pair of hard refs.

Cheers,
Erik

In reply to Re: Re: Getting a variable name from a reference by erikharrison
in thread Getting a variable name from a reference by Fool on the Hill

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