Prototypes alter parsing rules, alter the context in which arguments are evaluated, and limit what operators appear as arguments. Since those are all compile-time effects, prototypes have no run-time effects.

This means that $subref->() doesn't care if the referenced sub has a prototype or not. 

sub foo(\@) { ...} 

foo(@a);              # Passes \@a

my $subref = \&foo;
$subref->(@a);        # Passes $a[0]

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In reply to Re: sub routine help by ikegami
in thread sub routine help by hnd

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