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...

In reply to Re: Syntax to call a subroutine from a reference by davidrw
in thread Syntax to call a subroutine from a reference by neilwatson

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