in reply to Re: Calling a hash member
in thread Calling a hash member

Your first snippet doesn't work. For example, the following doesn't print 1 and 2. 
{ package ProjectA; { my $i; sub prepare { print ++$i, "\n"; } } }

my $project = 'ProjectA';
my %ThingsToDo = (firstTask => \{$project->prepare});

$ThingsToDo{firstTask};
$ThingsToDo{firstTask};

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Re^3: Calling a hash member
by geekphilosopher (Friar) on Dec 14, 2006 at 01:46 UTC

    Hrm, good catch. Do you happen to know why this doesn't work? Does the coderef not maintain the closure or something?

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      • my $r = \{expr};

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        is the same thing as 

        my @rv = expr;
my %h = @rv;
my $hr = \%h;
my $r = \$hr;

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        There's no coderef anywhere in there. That sets the value at key firstTask to a reference to a reference to a hash which was initialized with the result of prepare.

      • Furthermore, 

        $ThingsToDo{firstTask};

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        is a no-op. Even if $ThingsToDo{firstTask} returned a code reference, there's nothing to cause the referenced code to get executed. To execute the code referenced by $ThingsToDo{firstTask}, use 

        &{$ThingsToDo{firstTask}}();

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        or better yet, 

        $ThingsToDo{firstTask}->();

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      • Finally, $project needs to be variable. It needs to be an argument.

      I think you meant 

      my %ThingsToDo = (
   firstTask => sub { my $pkg = shift; $pkg->prepare(@_) },
);

...

$ThingsToDo{firstTask}->($project);

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      But that's much more complex than required, and would involve lots of redundancy (sub { my $pkg = shift; $pkg->XXX(@_) }.

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[d/l]
[select]
        Thanks a lot - I figured this was a good opportunity for a code ref, but I don't actually have much experience using them. Next time I'll have a go at a few sample programs before I offer (dubious) advice.
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