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
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.
so, my question is, what the scope of variables, defined in C and
C conditions?
in perl5:
my $a="first\n";
if (my $a="second\n") {print $a}
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