$Parameter is not a hash element it is an argument to the procedure reference stored in the hash.

I know. I didn't use ->{}. I used ->() which calls the referenced sub.

I'm sorry but that doesn't work either

Yes it does. 

#!/usr/bin/perl
use strict;
use warnings;

sub Procedure_Name_1
   {
   my($Parm) = shift;
   print($Parm);
   }

my %Procedures;
$Procedures{'ProcName1'} = \&Procedure_Name_1;

my $Procedure = 'ProcName1';
my $Parameter = 'Some Value';
$Procedures{$Procedure}->($Parameter);

[download]
>perl 730516.pl
Some Value

[download]

In reply to Re^7: How to call a sub reff from a hash by ikegami
in thread How to call a sub ref from a hash by NateTut

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