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Re^2: Lexical %+ %- and more?

by blazar (Canon)
on Oct 22, 2008 at 13:38 UTC ( #718748=note: print w/replies, xml ) Need Help??

in reply to Re: Lexical %+ %- and more?
in thread Lexical %+ %- and more?

What should that print?

I personally believe:

--- w: bar --- w: - bar - bar

The same "variable" is used and thus it is natural for it to be clobbered: if I didn't want, then I would have used a different one, especially since "now" it is so easy, whereas it wouldn't be an option were it only for numbered captures.

What if the last match was '--' =~ /(?<\w>\w+)/?

I beg your pardon, but... I don't see the difference! Maybe I'm just tired...

What if %+ is lexical, but %- isn't?

Well, they should behave independently, although of course this would be very inconsistent if one need both. (But I bet some hacker would find a cool way to exploit it for something weird and insane! ;)

And if lexical %- and %+ works as you want, should this work as well?
'foo' =~ /(?<w>\w+)/ && 'foofoo' =~ /\g{w}\g{w}/;

I don't see any reason why it shouldn't.

But that begs the question, what about:
'oo' =~ /(?<w>\w+)/ && 'oo' =~ /\g{w}\g{w}/;

Well, this should plainly fail. I think you're asking me what should be of %+ and %- after this, right? Well: no named captures are attempted in the second match, so they should stay like:

%+ = ( w => 'oo'); %- = ( w => ['oo']);

But if it were

'oo' =~ /(?<w>\w+)/ && 'oo' =~ /(?<w>\g{w}\g{w})/; # Which I *think* +is possible!

then they would become

%+ = ( w => undef); # or not existing at all? I'm half hearted... %- = ( w => []);
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