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Re: qr stringification: why are xism always present? I'm worried about backward compatibility

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August 2, 2010 01:26
Re: qr stringification: why are xism always present? I'm worried about backward compatibility
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On 2 August 2010 00:54, karl williamson <> wrote:
> Ben Morrow wrote:
>> Quoth (karl williamson):
>>> Ben Morrow wrote:
>>>> Quoth (karl williamson):
>>>>> But there must be some reason that xism are always shown in the
>>>>> stringification, whether they are in effect or not in effect.  But I can't
>>>>> think what it might be.  It would have been simpler for the original design
>>>>> to only output them in the stringification when different from the default.
>>>>>  So there must be a reason why the plus or minus of them is always output.
>>>>>  Does that reason apply to these new modifiers?
>>>> The reason is so that interpolation under different flags works
>>>> properly. Currently, if I have
>>>>    my $rx = qr/f./s;
>>>> is stringifies to (?s-xim:f.), which means that when I interpolate it
>>>> into a pattern with different flags
>>>>    "AFx" =~ /a$rx/i;
>>>> the /i is 'cancelled' for the duration of the interpolation (and /s
>>>> activated). If it were changed to stringify to (?s:f.) only, the match
>>>> above would change from failing to passing.
>>> 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.
>> Um, it's not what actually happens :)?
>>    ~/src/perl% perl -E'say "F" =~ /(?s:f)/i'
>>    1
>>    ~/src/perl% perl -E'say "F" =~ /(?s-ixm:f)/i'
>>    ~/src/perl%
>> Complete absence of a modifier means 'leave this modifier as it was
>> outside this group'. I don't think that can (or should) be changed now.
>> As Avar said, this could potentially be changed for the simple case of
>> direct interpolation, by looking up the flags on the interpolated regex
>> directly. However, a lot of code assumes it can stringify a qr//, write
>> it to a file/pass it across the network/whatever, and re-qr// to get an
>> equivalent pattern back.
>> My alternative suggestion was to introduce a new grouping construct,
>> which I tentatively called (?~sixm:) (I don't much like that, but there
>> aren't many alternatives at this point), which *does* do what you
>> expect; and use that for stringification instead. That way we change the
>> stringification once, now, and then never again.
>> Ben
> I think I like your idea.  Let me make sure I understand.  A tilde
> immediately following '(?' means that the clustering group is to have the
> default modifiers, subject to any flags that come after it.  Thus (?~:abc)
> means all default flags.  (?~x:abc) means the defaults except use the x
> modifier, and (?~-x:abc) means the defaults without the x modifier, which is
> a somewhat more complicated way of writing the equivalent (?~:abc).
> regex stringification could change to output just the non-defaults, or we
> could leave them as-is, and just add the tilde.  Since we would already be
> breaking compatibility, I would think we might as well remove them.
> A tilde is not the ideal choice because it can easily be confused with a
> minus sign.  I don't know if there are other characters that could be used
> instead.

Can you paint a picture for me as to how you think this is useful and
would work in practice? What problem it solves?

Right now I would vote against this idea in terms of minimal benefit
for long term costs, so maybe I don't understand the point properly.

Top of my mind questions:

How do you define "the default modifiers"? Assume we have them, and
they end up user configurable, what happens when a regex is compiled
in one context and the interpolated in another?

What is the difference between /(?~i:foo)/ and /(?i)foo/


perl -Mre=debug -e "/just|another|perl|hacker/"

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