From ilya@math.ohio-state.edu

$foo=1;$^W = 1;
sub test1 {my $foo = 2; return sub { eval shift } }
print test1()->('$foo');

P.S. See the *scope* where eval '' is executed? This *scope* has a
lexical $foo inherited from the enclosing lexical scope. Thus
the call test1()->('$foo') will return this lexical $foo.

 However\, since the "subsubroutine" of test1 is not a closure\, it
 does not \*preserve\* the reference to this $foo\.  Hence the
 refcount of $foo decreases when test1 returns\, which effectively
 makes $foo stale=undefined\.

 Putting $foo into the subsubroutine will make it a closure\, thus
 $foo will be 108 during the call\.

Thank you for the explanation of what happens. I now understand how it
is that test1's return value returns undef -- someone else had
explained it, but I think your explanation is clearer. I still think
test1 *should* return something different. But that's a topic for p5p.

This cannot be done. But I Cc perlbug​:

Here is what *can* be done. When evalling, internal (sub {}) scope
does not contain $foo. So the search for $foo continues up-scope.
Scope of test1 is tested, and $foo is found there.

POSSIBLE ENHANCEMENT​: given that the lexical was found inside eval '',
that the compilation of test1 is completed already, and that test1 is
not active any more, give a warning

  Access to stale lexical of enclosing scope from eval ''

[Variant​: additionally to testing that test1 is not active, test that
"the `active' position inside test1" is in the scope of $foo.]

Ilya

From @gbarr

This is still present in 5.8.0

From @iabyn

(Just reviewing old Perl bug reports)

The current development version of Perl (which will eventually be
released as 5.10.0) contains a number of lexical/closure fixes
and enhancements, which among other things, now cause this original
test code​:

  $foo=1;$^W = 1;
  sub test1 {my $foo = 2; return sub { eval shift } }
  print test1()->('$foo');

to emit the run-time warning​:

  Variable "$foo" is not available at (eval 1) line 2.

since at the time of the eval, the lexical $foo is not active.
Note that it does *not* instead use the value of the global $foo.

(This warning is the equivalent of the 'stale' warning that Ilya
originally suggested).

I am therefore making this report a resolved.

Regards,

Dave M.

@iabyn - Status changed from 'open' to 'resolved'