Re^7: How to call a sub reff from a hash

$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);
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>perl 730516.pl
Some Value
[download]

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Re^8: How to call a sub reff from a hash
by NateTut (Deacon) on Dec 16, 2008 at 04:49 UTC

I apologize you are right of course, I must have copied wrong before. The real problem here is that I'm not understanding the syntax. Thanks for all the help.

[reply]
[d/l]

Re^9: How to call a sub reff from a hash

by ikegami (Patriarch) on Dec 16, 2008 at 04:52 UTC

I'm not understanding the syntax

This Can be written more clearly as

${$aref}[$i] $aref->[$i]

${$href}{$k} $href->{$k}

&{$cref}($arg) $cref->($arg)

In this case, the code ref is stored in a hash, so $cref is a hash lookup.

my $cref = $hash{$k}; $cref->($arg)

    |
    v

( $hash{$k} )->($arg);

    |
    v

$hash{$k}->($arg);
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