There are a couple of things happening here.

1) Inside of the closure the var $hashref is re-my'd and thus is a completely new variable, therefore no reference to %monks is transmitted outside the closure. This means that $hashref is undef outside the closure.

2) If you remove the second my and still initialize $hashref to point to %monks then the memory will not be cleaned up because there still exists a variable that has a reference pointing to it so its reference count is not zero but the memory that was previously refered to as %monks is no longer accessible by that name but is accessible by %$hash_ref. Now, if you make $hash_ref = undef; then the final reference to the memory previously labeled %monks will finally go away thus making the reference count to that piece of memory zero and Perl will clean it up if it feels it needs to, otherwise that memory will just go back into the pool to be reused.

At least that is the way I understand it. Those with more internals experience may prove me wrong. =)

Cheers!
--habit

In reply to Re: Memory leaks and reference counting within Perl. by habit_forming
in thread Memory leaks and reference counting within Perl. by DigitalKitty

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