Here you are strongly suggesting a symblic reference. As for the code, the part before && obviously evaluate to false, so that exists after && will be evaluated, but no auto-vivfication here, as $foo{bar} has been created by you earlier

Actually, a) the ref failing will immediately lead to the entire expression failing because he is using &&, not ||. b) If it got executed, strict would cause the exists to fail with a refs stricture violation.

And the latter is exactly what he's talking about protecting against by using ref.

Makeshifts last the longest.