Lexicals are hidden from any subroutine called from their scope. This is true even if the same subroutine is called from itself or elsewhere—each instance of the subroutine gets its own "scratchpad" of lexical variables.

Update 2: One might argue that the anomalous behavior of $x is consistent with the above dictum, because the subroutine foo is not being called from scope of $x. However, when the Perl wise ones write that lexicals are hidden from any subroutine called from their scope, it means even if they are called from their scope.

If a lexical is hidden from subroutines called from inside its own scope, then how much more so must it be hidden from subroutines called from outside its own scope.

As Abigail has taught, conditionally declaring lexicals for use in subroutines is an example of "an ox that is known to gore," i.e., its behavior is going to be ambiguous. Even though it is legal to release such an ox and let it run around and possibly damage a neighbor's code, let us instead "build a fence around the Law" and only declare lexical variables in subroutines unconditionally, outside of conditional branching such as 'my $x if 0'.

(previous updates effaced)

Well, technically what is happening is not violating Camel III. See, it *does* get its own scratchpad, which can be quickly checked by:

    #!/opt/perl/bin/perl -w
    use strict;
    sub foo;
    sub foo {
        return unless $_ [0];
        my $x if 0;
        print ++ $x, " ";
        foo $_ [0] - 1;
    }
    
    foo 1; print "\n";
    foo 2; print "\n";
    foo 3; print "\n";
    foo 4; print "\n";
    __END__
    1 
    2 1 
    3 2 1 
    4 3 2 1
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Scary, isn't? When the subroutine recurses, it notices there is still a reference to $x and hence it will create a new scratchpad. But when there is no reference, it will reuse an old scratchpad....

I do agree that using my in this way doesn't tend to lead to well understood code. I wouldn't go as far as to say it's never useful, but such cases will be very rare and it's not a technique I would teach in my classes.

-- Abigail

Come back Abigail(II). This place needs you.

Have you tried this?

#!/usr/bin/perl
use strict;
    sub foo;
    sub foo {
        return unless $_ [0];
        my $x if undef;
        print ++ $x, " ";
        foo $_ [0] - 1;
    }

    foo 29; print "\n";
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It prints:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Hum...