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Just think of it this way: when a function closes over a lexical variable, that instance of the variable (and its associated storage location) won't be freed, as it normally would when the variable goes out of scope — i.e. it continues to remain accessible to the function.

Note that you can (in theory) have more than one closure sharing the same variable instance (kind of their private global variable). Extending the Anonymous Monk's code:

#!/usr/bin/perl sub make_counter_pair { my $start = shift; warn "lexical variable ".\$start." created"; return ( sub { warn "accessing ".\$start; $start++ }, sub { warn "accessing ".\$start; $start++ } ) } my ($c1, $c2) = make_counter_pair(10); my ($c3, $c4) = make_counter_pair(3); print $c1->() . "\n"; print $c2->() . "\n"; print $c1->() . "\n"; print $c2->() . "\n"; print $c3->() . "\n"; print $c4->() . "\n"; print $c3->() . "\n"; print $c4->() . "\n";

would give

lexical variable SCALAR(0x604ff0) created at ./837669.pl line 5. lexical variable SCALAR(0x6323f0) created at ./837669.pl line 5. accessing SCALAR(0x604ff0) at ./837669.pl line 8. 10 accessing SCALAR(0x604ff0) at ./837669.pl line 12. 11 accessing SCALAR(0x604ff0) at ./837669.pl line 8. 12 accessing SCALAR(0x604ff0) at ./837669.pl line 12. 13 accessing SCALAR(0x6323f0) at ./837669.pl line 8. 3 accessing SCALAR(0x6323f0) at ./837669.pl line 12. 4 accessing SCALAR(0x6323f0) at ./837669.pl line 8. 5 accessing SCALAR(0x6323f0) at ./837669.pl line 12. 6

As you can see by looking at the addresses, each pair of closures shares one variable, so the respective value is incremented every time one of the two counter functions is called.


In reply to Re: Closure Explanation by almut
in thread Closure Explanation by bichonfrise74

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