According to Larry Wall:
> I don't claim to follow all this talk about "stores"
Think about tied values. When does STORE get called, precisely, on a
tied target of s///? It's good to be explicit about this, down at the
C API level, just so we know what to optimize for. The final answer
is proba
On Wed, Sep 08, 2004 at 11:00:54PM -0700, Steve Fink wrote:
: I vote for leaving all of these sorts of cases undefined. Well,
: partially defined -- I'd rather we didn't allow ($a = "aaa") =~ s/a/b/g
: to turn $a into "gawrsh". At the very least, define the exact number of
: output and stores for "
On Sep-09, [EMAIL PROTECTED] wrote:
> On Wed, 8 Sep 2004, Chip Salzenberg wrote:
>
> > According to [EMAIL PROTECTED]:
> > > So how many stores do we expect for
> > >($a = "xxx") =~ s/a/b/g
> > > and which of the possible answers would be more useful?
> >
> > I think it depends on C<($a = "aaa
On Wed, 8 Sep 2004, Chip Salzenberg wrote:
> According to [EMAIL PROTECTED]:
> > So how many stores do we expect for
> >($a = "xxx") =~ s/a/b/g
> > and which of the possible answers would be more useful?
>
> I think it depends on C<($a = "aaa") =~ s/a/b/g>.
I would agree with you in general,
According to [EMAIL PROTECTED]:
> So how many stores do we expect for
>($a = "xxx") =~ s/a/b/g
> and which of the possible answers would be more useful?
I think it depends on C<($a = "aaa") =~ s/a/b/g>.
* If the s/// operator stores once after all substitutions,
then having it alway store
On Tue, 7 Sep 2004, Leopold Toetsch wrote:
> > [*] Unless it's a _feature_ that given tied $a,
> >($a = "aaa") =~ s/a/b/g
> > would call STORE four times ("aaa", "baa", "bba", "bbb").
>
> I'd expect two stores here. One for the initial setting of the value and
> one for the final result
At 10:59 PM -0400 9/6/04, Chip Salzenberg wrote:
Just across the hall from m// is s/// ...
Considering the semantics of m// and especially s/// at the user
level, we'll probably[*] want to take snapshots of dynamic strings
(think P5's "FETCH" or overload '""'), and apply all the pattern
operations
Chip Salzenberg <[EMAIL PROTECTED]> wrote:
> For Topaz, Scalar's interface included a function that would basically
> open the Scalar's hood, giving you a Buffer you could manipulate; then
> when you were done working with the Buffer, its modifications (if any)
> were propagated back down into the
Just across the hall from m// is s/// ...
Considering the semantics of m// and especially s/// at the user
level, we'll probably[*] want to take snapshots of dynamic strings
(think P5's "FETCH" or overload '""'), and apply all the pattern
operations to that snapshot. *However*, in the usual case
