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Postings from August 2010
Ævar Arnfjörð Bjarmason wrote:
> On Sun, Aug 1, 2010 at 19:44, karl williamson <public@khwilliamson.com> wrote:
>
>> I still don't understand. I had considered interpolation, and it still
>> seems to me that the absence of a modifier means use the underlying default
>> for the duration of that interpolation, no matter what the outer regex says.
>> I don't see a flaw in that thinking.
>
> It's not an issue if you have a regexp object. But consider this:
>
> $ perl -E 'my $s = "foo"; my $o = qr/foo/; my $so = "$o"; my $n =
> qr/ $s ${o} blah ${so} /i; say $n'
> (?i-xsm: foo (?-xism:foo) blah (?-xism:foo) )
>
> Even if we didn't dump the default modifiers we could detect the "$o"
> case since it's a regexp object, but we wouldn't detect "$so" since
> it's a stringified "$o".
>
I think I understand now, and perhaps it was so obvious to you guys that
it didn't get said in a way I can understand. I get the sense that some
of the stuff I say about Unicode is so obvious to me that I don't
realize it isn't so clear to others without as much background in the topic.
What I wasn't realizing is that a clustering parentheses group (?:...)
inherits the outer modifiers, unless overridden. Thus in
/ (?:foo) blah/i
the foo is matched case insensitively. If it isn't to be matched with
/i, the modifiers must appear as in
/ (?-i:foo) blah/i
Therefore the modifiers must be present always when interpolating. And
Ben's suggestion makes sense if there is a substantial body of code out
there that relies on stringification never changing.
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