Nope. Each coderef (set of coderefs in your example - it's the time that the lexical is in scope that matters) has it's own copy.

sub make_closures
{
  my $value = shift;
  return sub { $value++ }, sub { print "$value\n" }, sub { $value-- };
}

my ($increment, $display, $decrement) = make_closures( 3 );
print "first\n";
$display->();
$increment->();
$display->();
$decrement->();
$display->();

my ($increment2, $display2, $decrement2) = make_closures( 9 );

print "second\n";
$display2->();
$increment2->();
$display2->();
$decrement2->();
$display2->();

print "first again\n";
$display->();
$increment->();
$display->();
$decrement->();
$display->();
[download]

Result:

first
3
4
3
second
9
10
9
first again
3
4
3

That's why it needs to be lexical.

I suspect we mean different things by the word "copy".

Because lexical variables live in scratchpads and because a closure carries around the scratchpad where it was created, I say that a closure does not make a new copy of the variable.

Because a new scratchpad is created on each scope entry and the closures returned bind to the new scratchpad, you say that a closure does copy the variable.

I don't think "copy" is the right word, but I understand what you're saying and the effect is indeed accurate.

If by "same" you mean it's the same $item it was during the life of the call to shoppingList(), even though it lives in a different scratchpad from that of all other calls to shoppingList(), then I understand what you meant. And you certainly understand what I meant. (is chromatic here the same as chromatic on slashdot? If so, I probably shouldn't argue at all.)



--Bob Niederman, http://bob-n.com