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Article <mailman.28.1729285790.4695.python-list@python.org>
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From: "Peter J. Holzer" <hjp-python@hjp.at>
Newsgroups: comp.lang.python
Subject: Re: Correct syntax for pathological re.search()
Date: Fri, 18 Oct 2024 23:09:41 +0200
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On 2024-10-12 08:51:57 -0400, Thomas Passin via Python-list wrote:
> On 10/12/2024 6:59 AM, Peter J. Holzer via Python-list wrote:
> > On 2024-10-11 17:13:07 -0400, AVI GROSS via Python-list wrote:
> > > Is there some utility function out there that can be called to show w=
hat the
> > > regular expression you typed in will look like by the time it is read=
y to be
> > > used?
> >=20
> > I assume that by "ready to be used" you mean the compiled form?
> >=20
> > No, there doesn't seem to be a way to dump that. You can
> >=20
> >      p =3D re.compile("\\\\sout{")
> >      print(p.pattern)
> >=20
> > but that just prints the input string, which you could do without
> > compiling it first.
>=20
> It prints the escaped version,

Did you mean the *un*escaped version? Well, yeah, that's what print
does.

> so you can see if you escaped the string as you intended. In this
> case, the print will display '\\sout{'.

print("\\\\sout{")
will do the same.

It seems to me that for any string s which is a valid regular expression
(i.e. re.compile doesn't throw an exception)

    assert re.compile(s).pattern =3D=3D s

holds.

So it doesn't give you anything you didn't already know.

As a trivial example, the regular expressions r"\\sout{" and r"\\sout\{"
are equivalent (the \ before the { is redundant). Yet
re.compile(s).pattern preserves the difference between the two strings.

        hp

--=20
   _  | Peter J. Holzer    | Story must make more sense than reality.
|_|_) |                    |
| |   | hjp@hjp.at         |    -- Charles Stross, "Creative writing
__/   | http://www.hjp.at/ |       challenge!"

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