Re^2: Debugger and lexicals

[Anonymous Monk]

No; there is no control or sandbox and the debugger, being written in perl, is part of perl.

Lexical variables are only accessible from within a lexical scope, because they only exist within a lexical scope.

You can use padwalker (y) to list lexicals in a higher scope.

I think jethro is right, and my $a probably gets optimized away

[Khen1950fx]

Some of what you say I can agree with. But "they only exist within a lexical scope"---and why's that? It's because lexical variables cannot access the symbol table. That's why I mentioned that there's a control "sandboxed" way from the debugger. Package variables can be accessed via the symbol table.

As for my $a getting optimized away, I disagree with that because I just don't see how it can be optimized away when it's uninitialized in the subroutine. Here's a little test that I ran:

#!/usr/bin/perl

use strict;
use warnings;

print mt2(55), "\n";

my $a = 23;

sub mt2 { 
    my $b = $_[0] * 2;
    return $a + $b;
}

Here's the corrected script:

#!/usr/bin/perl

use strict;
use warnings;

print mt2(55), "\n";

sub mt2 { 
    my $a = 23;
    my $b = $_[0] * 2;
    return $a + $b;
}

This one works in the debugger. Thanks for the input.

[Anonymous Monk]

and why's that? It's because lexical variables cannot access the symbol table.

No, its because they're initialized at runtime and stored in a pad/Scratchpads, which are attached to subroutines ... http://search.cpan.org/dist/illguts/

As for my $a getting optimized away, I disagree with that because I just don't see how it can be optimized away when it's uninitialized in the subroutine. Here's a little test that I ran:

Your example is different.

In your example, $a is initialized after the call to mt2.

And you never remove a reference to $a, so it never gets optimized away.

Here is what you intended :)

$a is defined

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The "problem" of $a being undefined, is because it is optimized away -- really because the sub which we're stepping into (ASDF) doesn't refer to it

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Here, a different sub refers to $a, so when we step into ASDF, $a is still undefined -- it is not attached to ASDF's Scratchpad

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You should consider not answering or prefacing your guesses (see To Answer, Or Not To Answer....), because there won't always be someone there to answer when Duty Calls

[dave_the_m]

Stricly speaking $a is not optimised away; it just has a short life span, as as I pointed out earlier.

$ cat /tmp/Foo.pm 
package Foo;
sub DESTROY { print "DESTROY\n" }
my $a = bless {};
print "executing Foo\n";

$ perl -I/tmp -e'use Foo; BEGIN { print "BEGIN\n" }'
executing Foo
DESTROY
BEGIN

Dave.