Re: Lexical %+ %- and more?

my %+;
my %-;
'foo' =~ /(?<w>\w+)/;
'bar' =~ /(?<w>\w+)/;
use YAML;
print Dump \%+;
print Dump \%-;
__END__
[download]

What should that print? What if the last match was '--' =~ /(?<\w>\w+)/? What if %+ is lexical, but %- isn't?

And if lexical %- and %+ works as you want, should this work as well?

'foo' =~ /(?<w>\w+)/ && 'foofoo' =~ /\g{w}\g{w}/;
[download]

But that begs the question, what about:

'oo' =~ /(?<w>\w+)/ && 'oo' =~ /\g{w}\g{w}/;
[download]

Comment on Re: Lexical %+ %- and more? Select or Download Code

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Re^2: Lexical %+ %- and more? by blazar (Canon) on Oct 22, 2008 at 13:38 UTC
What should that print? I personally believe: `--- w: bar --- w: - bar - bar` [download] 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}/;` [download] 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}/;` [download] 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']);` [download] But if it were `'oo' =~ /(?<w>\w+)/ && 'oo' =~ /(?<w>\g{w}\g{w})/; # Which I think +is possible!` [download] then they would become `%+ = ( w => undef); # or not existing at all? I'm half hearted... %- = ( w => []);` [download] `--` If you can't understand the incipit, then please check the IPB Campaign.	[reply] [d/l] [select]

Replies are listed 'Best First'.

Re^2: Lexical %+ %- and more?
by blazar (Canon) on Oct 22, 2008 at 13:38 UTC

What should that print?

I personally believe:

---
w: bar
---
w:
  - bar
  - bar
[download]

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}/;
[download]

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}/;
[download]

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']);
[download]

But if it were

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

then they would become

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

--
If you can't understand the incipit, then please check the IPB Campaign.

[reply]
[d/l]
[select]