Can you guess what the following snippet prints?

for (1..4) {
   my $x;
   print ++$x, " ", \$x, "\n";
   sub { $y = $x };
}
[download]

And this?

for (1..4) {
   my $x;
   print ++$x, " ", \$x, "\n";
   sub bob { $y = $x };
}
[download]

And now the answer: In the first case there is no closure, so only one $x is created, but in the second case there is a closure so 2 different $xs are created as you can see from the output.

for example:

1 SCALAR(0x8160300)
1 SCALAR(0x8160300)
1 SCALAR(0x8160300)
1 SCALAR(0x8160300)

1 SCALAR(0x81649e8)
1 SCALAR(0x814cd38)
1 SCALAR(0x814cd38)
1 SCALAR(0x814cd38)