in reply to Syntax to call a subroutine from a reference

you almost had it: 
$x->{'result'} = &{$x->{'subname'}}();

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Update: See perlref, especially example 2 in the "Using References" section.

Update: As herveus notes, your original way of $x->{'result'} = $x->{'subname'}->(); does work as well...

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Re^2: Syntax to call a subroutine from a reference
by herveus (Prior) on Sep 21, 2005 at 18:17 UTC
    Howdy!

    $x->{subname}->() (as seen in the original code) also works fine. In fact, I would tend to prefer that over using the ampersand sigil.

    yours,
    Michael
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