calling from outside the closure a sub defined in a closure

spx2 has asked for the wisdom of the Perl Monks concerning the following question:

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Re: calling from outside the closure a sub defined in a closure by Anno (Deacon) on Sep 19, 2007 at 09:48 UTC
Your code doesn't show a closure, just a bare block that defines a coderef (which might itself be a closure) in its lexical scope. If you want to access that outside the block, you must find a way to pass the coderef outside. A sub could return it, probably among other things. This looks like a somewhat convoluted design. Anno	[reply]
Re: calling from outside the closure a sub defined in a closure by derby (Abbot) on Sep 19, 2007 at 12:17 UTC
Anno was right, that's not a closure. Here's the typical example of a closure: `#!/usr/bin/perl use strict; use warnings; my $plus_5 = adder( 5 ); my $plus_10 = adder( 10 ); print $plus_5->(2), "\n"; print $plus_10->(3), "\n"; sub adder { my $base = shift; return sub { $base + shift; } }` [download] -derby	[reply] [d/l]
Re: calling from outside the closure a sub defined in a closure by moritz (Cardinal) on Sep 19, 2007 at 09:30 UTC
If you want to use it outside the sub, you have to return it to the caller (or use a global/package variable, but you should avoid that). You can call it wherever you have a reference to it. Perl 6 in German	[reply]
Re: calling from outside the closure a sub defined in a closure by siva kumar (Pilgrim) on Sep 19, 2007 at 09:30 UTC
You can't. You can use if the subroutine in local scope. `{ local $subRef = sub { $var = shift; print "hello monks --- [$var] \n"; }; &callFun(); } sub callFun { &$subRef(20); }` [download] OR `{ $subRef = sub { $var = shift; print "hello monks --- [$var] \n"; }; } &$subRef;` [download]	[reply] [d/l] [select]