#!/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);
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Because you have no scalar $Procedures declared (you have a hash %Procedures, but that's something completely different).

The cake is a lie.
The cake is a lie.
The cake is a lie.

Yes, but I'm trying to access an element of that hash.

For that you want

&{$Procedures{$Procedure}}($Parameter);
# or
$Procedures{$Procedure}->($Parameter);
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And yeah, the syntax may be confusing, but that's just how references work.
"What should it profit a man, if he should win a flame war, yet lose his cool?"

Taking your code and then switching the last line like he said makes it work perfectly.

#!/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);
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I don't understand why you ignored his advice but i would recommend trying it as written in the future.

BTW you could read the last line as "get the value of $Procedures{$Procedure} and then run it as a coderef ->( passing it $Parameter)"


___________
Eric Hodges