Re: Lost anonymous subs

No, @arr is still accessible:

my @arr = (9) x 1000_000;
my $code = sub { @arr };
undef $code;
print(scalar(@arr), $/);  # 1000000
[download]

You might mean

my $code;
{
   my @arr = (9) x 1000_000;
   $code = sub { @arr };
}
# @arr kept alive by the sub
undef $code;
# @arr freed
print(scalar(@arr), $/);  # error under strict.
[download]

@arr would be freed at the undef. The memory is released to perl, but maybe not to the OS. There has been discussion on this in the past.

Nit: perl uses reference counting, not garbage collection.

Comment on Re: Lost anonymous subs Select or Download Code

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Re^2: Lost anonymous subs by kappa (Chaplain) on Dec 09, 2004 at 16:02 UTC
I have written almost exactly the same piece of code in the reply to diotalevi up there because did not spot your comment. Yes, that helped me to understand things. I somehow got thinking that closures copy referenced lexicals when created. That was my error. This is the case when copying really takes place: `my $Big = 'X' x 1000000; sub big { my $big = $Big; return sub { $big }; } my @crefs; push @crefs, big() for 1..10;` [download] --kap	[reply] [d/l]
Re^3: Lost anonymous subs by diotalevi (Canon) on Dec 09, 2004 at 17:21 UTC
Well yes, that's because the = operator copies stuff. There's nothing special about that.	[reply]
Re^4: Lost anonymous subs by kappa (Chaplain) on Dec 10, 2004 at 12:07 UTC
`my @abc; for my $m qw/a b c/ { push @abc, sub { $m }; } print $_->() . "\n" for @abc;` [download] No assignments and still, those subs each has its own `$m`. Or does the `for my $m` creates a new variable on each iteration? --kap	[reply] [d/l] [select]
Re^5: Lost anonymous subs by diotalevi (Canon) on Dec 10, 2004 at 15:02 UTC
Re^6: Lost anonymous subs by kappa (Chaplain) on Dec 10, 2004 at 15:27 UTC
Some notes below your chosen depth have not been shown here
Re^3: Lost anonymous subs by ysth (Canon) on Dec 10, 2004 at 11:50 UTC
Not exactly copy, more just steal. Usually lexicals hang around for the next time they go into scope (though they get reset to be empty/undef, they still hold memory). If you take a reference to one (including the case of a closure) and hold the reference until the lexical goes out of scope, the lexical has to get reallocated, since the remaining reference keeps the original.	[reply]
Re^2: Lost anonymous subs by chb (Deacon) on Dec 10, 2004 at 08:55 UTC
Nit Nit: perl uses reference counting to see what can be safely garbage collected AFAIK.	[reply]