> Should properties interpolate in regular expressions? (and/or strings)
Do you mean property look-ups via the pseudo-method syntax?
In that case, yes they should
> I don't suppose they should, because we don't expect subroutines to.
> (if $foo =~ /bar($baz,$quux)/;? Urgh, maybe we need m//e)
Err. I *would* expect sub call iterpolation in regexes, since they will
interpolate in qq{...} contexts, and that's what a regex basically is.
But what you showed is not the syntax for it:
/bar($baz,$quux)/; # match bar then match-and capture
# matches of interpolated value of $baz,
# a comma, interpolated value of $quux
/&bar($baz,$quux)/; # match interpolated value of call to
# bar with two args
> What should $foo = (1,2,3) do now? Should it be the same as what
> $foo = [1,2,3]; did in Perl 6? (This is assuming that $foo=@INC does what
> $foo = \@INC; does now.) Putting it another way: does a list in scalar
> context turn into a reference, or is it just arrays that do that?
Just arrays, I believe.
> Currently I have:
>
> % ./perl -l
> printf "This is Perl version %vd\n", $^V;
> %foo = (test=>"ok 1", test2=>"ok 3");
> print %foo{test};
Yes
> print "ok 2" if ref ($a=%foo);
Yes. Though C<ref($a=%foo) eq 'HASH'> would be a better test
> print $a->{test2};
die "Unexpected > after subtraction operation. Did you mean $a.{test2}???"
> print "ok 4" if ref @INC;
Yes
> print "ok 5" unless ref ($a=(1,2,3))'
No. Equivalent to ref($a=3), I believe
> Does that look right?
60% right, at least. ;-)
> > print $a->{test2};
>
> Oh, hrm. Shouldn't it be $a{test2}?
Yes. Or $a.{test}
> That works too, at any rate.
> Does that mean that arrow between variable and subscript is optional,
<Zuul voice>There is no arrow. Only dot.</Zuul voice>
And yes, it's optional anywhere the dot acts like a /\b/ boundary:
$ref.[1] can be $ref[1]
$ref.{a} can be $ref{a}
$ref.(@args) can be $ref(@args)
$ref.meth() CAN'T be $refmeth()
Disclaimer: Rules #1 and #2 apply to all of the above.
Damian
