On Tue, 29 Jun 2004 10:52:31 -0400, Jonadab The Unsightly One
<[EMAIL PROTECTED]> wrote:
People who think in terms of "statements" often get mixed up when they
put complex expressions in void context, expecting them to be treated
as statements. print(2+3)*7; is another example. Perl doesn't ha
John Williams <[EMAIL PROTECTED]> writes:
> $b = 'a';
> my $b ='b' , print "$b\n";
> print "$b\n";
>
> Which seems to show that the "my $b" doesn't actually come into
> scope until the end of the statement in which it is defined.
The comma operator doesn't guarantee order of operation becau
Luke Palmer writes:
> Alexey Trofimenko writes:
> > of course, I just mutter.. new C is very good, and in special
> > cases, when simple incrementing-decrementing isn't what I want, I can
> > write my own iterator (btw, in which apocalypse I can find how to
> > write iterators in perl6?) with m
Alexey Trofimenko writes:
> of course, I just mutter.. new C is very good, and in special
> cases, when simple incrementing-decrementing isn't what I want, I can
> write my own iterator (btw, in which apocalypse I can find how to
> write iterators in perl6?) with my own custom very special incr
On Mon, 28 Jun 2004 06:42:47 -0700, David Storrs <[EMAIL PROTECTED]>
wrote:
On Sun, Jun 27, 2004 at 03:16:11PM -0600, Luke Palmer wrote:
But anyway, if you still want to be old school about it, then you'll end
up not caring about the scope of your $i. Really you won't. And you'll
be happy that
On Mon, Jun 28, 2004 at 11:10:03AM -0600, John Williams wrote:
> On Sun, 27 Jun 2004, Luke Palmer wrote:
>
> > Alexey Trofimenko writes:
> > > AFAIR, I've seen in some Apocalypse that lexical scope boundaries will be
> > > the same as boundaries of block, in which lexical variable was defined.
>
On Sun, 27 Jun 2004, Luke Palmer wrote:
> Alexey Trofimenko writes:
> > AFAIR, I've seen in some Apocalypse that lexical scope boundaries will be
> > the same as boundaries of block, in which lexical variable was defined.
>
> Yep. Except in the case of routine parameters, but that's nothing new.
On Sun, Jun 27, 2004 at 03:16:11PM -0600, Luke Palmer wrote:
>
> But anyway, if you still want to be old school about it, then you'll end
> up not caring about the scope of your $i. Really you won't. And you'll
> be happy that it was kept around for you once you decide you want to
> know the val
Alexey Trofimenko writes:
> AFAIR, I've seen in some Apocalypse that lexical scope boundaries will be
> the same as boundaries of block, in which lexical variable was defined.
Yep. Except in the case of routine parameters, but that's nothing new.
>
> so, my question is, what the scope of var
