in reply to Re: redeclaring variables with 'my'
in thread redeclaring variables with 'my'

It's not re-allocating memory each time through the loop. You can test that by doing this: 
    my $count = 1;
    while (my $x = $count) {
        print \$x, "\n";
        last if ++$count > 5;
    }

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If you run that, you should get something like: 
    SCALAR(0x80e4aa8)
    SCALAR(0x80e4aa8)
    SCALAR(0x80e4aa8)
    SCALAR(0x80e4aa8)
    SCALAR(0x80e4aa8)

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Each time through the loop $x is at the same memory location. So it's not re-allocating memory.

As tilly mentions in his post, Perl does have to do *some* extra things--but re-allocating memory isn't one of them. :)

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RE: RE: Re: redeclaring variables with 'my'
by Fastolfe (Vicar) on Nov 11, 2000 at 02:39 UTC
    If it were tightly de-allocating and re-allocating, is it not unreasonable to expect that they might end up at the same address? Not that I disagree with you; I'm hardly the expert. I rather think that what you say is actually the case.

    But yah, I think tilly's hit it on the head.

[reply]
      It would be reasonable, but it doesn't work that way. In fact here is a piece of code that triggers an illustrative bug in how it does work: 
      recurse_lex(1);
print "\n";
recurse_lex(2);
print "\n";
recurse_lex(3);
print "\n";
recurse_lex(4);
print "\n";

sub recurse_lex {
  my $i = shift;
  my @array if 0;
  unshift @array, $i;
  print @array, "\n";
  $i--;
  if ($i) {
    recurse_lex($i);
  }
}

      [download]
      Try that. It does something weird. And here are the details of what does happen.
[reply]
[d/l]
        This makes sense. I wasn't disagreeing, though. :)
[reply]

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