Try this:
#!/usr/bin/perl
foo();
bar();
for (1..3) {
    my $x;
    sub foo { print ++$x, "f\n" }
    sub bar { print ++$x, "b\n" }
    print ++$x, "m\n";
}
foo();
bar();
resulting in 
1f
2b
3m
1m
1m
4f
5b
foo and bar hold the first incarnation of x in the loop. The second and third incarnation of my $x is initialised with 0 so you get the 1m 1m lines in the midd.

In reply to Re: Trying to understand closures by Anonymous Monk
in thread Trying to understand closures by shine22vn

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