How do closures in Perl work?

Question

Newbie in Perl again here, trying to understand closure in Perl.

So here\'s an example of code which I don\'t understand:

sub make_saying  {         


        

Answer 1

Every time you call the subroutine 'make_saying', it:

1. creates a DIFFERENT closure

2. assigns the received parameter to the scalar '$salute'

3. defines (creates but not execute) an inner anonymous subroutine: That is the reason why at that moment nothing is assigned to the scalar $target nor is the statement print "$salute, $target!\n"; executed .

4. finally the subroutine 'make_saying' returns a reference to the inner anonymous subroutine, that reference becomes the only way to call the (specific) anonymous subroutine.

Ever time you call each anonymous subroutine, it:

1. assign the received parameter to the scalar $target

2. sees also the scalar $salute that will have the value assigned at the moment in which was created the anonymous subroutine (when was called its parent subroutine make_saying

3. finally executes the statement print "$salute, $target!\n";