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

In Perl, scalar variables cannot hold subroutines directly, they can only hold references. This is very much like scalars cannot hold arrays or hashes, only arrayrefs or hashrefs.

The sub { ... } evaluates to a coderef, so you can directly assign it to a scalar variable. If you want to assign a named function (e.g. foo), you have to obtain the reference like \&foo.

You can call coderefs like $code->(@args) or &$code(@args).

The code

$f = \make_saying("Howdy")

evaluates make_saying("Howdy"), and takes a reference to the returned value. So you get a reference that points to a coderef, not a coderef itself.

Therefore, it can't be called like &$f("world"), you need to dereference one extra level: &$$f("world").

---

A closure is a function that is bound to a certain environment.

The environment consists of all currently visible variables, so a closure always remembers this scope. In the code

my $x;
sub foo {
  my $y;
  return sub { "$x, $y" };
}

foo is a closure over $x, as the outer environment consists of $x. The inner sub is a closure over $x and $y.

Each time foo is executed, we get a new $y and therefore a new closure. Each time it is called, a different closure is returned.

When we execute make_saying("Howdy"), the $salute variable is set to Howdy. The returned closure remembers this scope.

When we execute it again with make_saying("Greetings"), the body of make_saying is evaluated again. The $salute is now set to Greetings, and the inner sub closes over this variable. This variable is separate from the previous $salute, which still exists, but isn't accessible except through the first closure.

The two greeters have closed over separate $salute variables. When they are executed, their respective $salute is still in scope, and they can access and modify the value.