in reply to Re: Can't localize lexical variable $var at...
in thread Can't localize lexical variable $var at...

you wrote:
In particular, a sub outside the scope cannot see that variable.

But subs inside that scope are able to see that! And I cleanly see a way how that lexically scoped variable could be stacked, just like global variable!

But yes, I'll better just use lexical variables and implement stacking if I really need.

Best wishes,
Courage, the Cowardly Dog

  • Comment on Re: Re: Can't localize lexical variable $var at...

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Re^3: Can't localize lexical variable $var at...
by Aristotle (Chancellor) on Jun 09, 2002 at 20:29 UTC
    No, it cannot. If you { my $var; some_sub(); }, then some_sub() cannot see $var. On the other hand, if you { local $var; some_sub(); }, then some_sub() can. (And that's why you shouldn't local unless you have a clear reason for it and not before you have signed a contract that you clearly understand that perl is not liable if you botch up at that point, and that you accept the consequences.)

    Makeshifts last the longest.

[reply]
[d/l]
[select]
      You did not understood what I meant. 
      {
  my $var=2;
  sub inc {return $var++}
  sub dec {return $var--}
}
print inc;
print dec;

      [download]
      Those subs see that $var which is not seen outside that block, and perl doc says this is a trick to have private static variables.

      Courage, the Cowardly Dog.

[reply]
[d/l]
        Now you are talking about closures. I do not see how that is relevant to localized globals?

        Makeshifts last the longest.

[reply]

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