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From: Andreas Leitgeb <avl@logic.at>
Newsgroups: comp.lang.java.programmer
Subject: Re: regex replaceAll?
Date: Fri, 2 Mar 2018 17:46:07 -0000 (UTC)
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Eric Douglas <e.d.programmer@gmail.com> wrote:
> out = in.replaceAll("(?<=\\G.*b.*)b", "_");

You might have noticed that the original RE (with spaces) uses
\s and \S, and that any single character can only *either* be
matched by \s *or* by \S. You can consider \S to be the opposite
of \s.

The opposite of b isn't ".", but rather "[^b]".

Note, that if "b" really should be a possibly longer string, 
then it's "opposite" will quickly become very ugly with
increasing length. That would be a case were you'd better
pick an alternative without REs.

So, by correcting the "." to "[^b]", you get:
out = in.replaceAll("(?<=\\G[^b]*b[^b]*)b", "_");
and to that you can now add the extra pair of non-
capturing parens and e.g. "{2}" for n=3 .

> The documentation on that says:
> (?<=X)	X, via zero-width positive lookbehind
>  (not sure what this did for the original expression here)

It makes sure that the "\\G[^b]*b[^b]*" is *not* part of the
thing to replace, yet required to exist before the actual match.

Your own approach of capturing the "prefix" with parens and
substing it into the replacement via $1 looks much cleaner
to me, btw.

> \G	The end of the previous match
>  (this is what I expect replaceAll to always do)

Well, nope.  if you do a "ababab".replaceAll("b","_"); you'd
want all "b"s replaced, even if not adjacent to the previous
one.

> \S	A non-whitespace character: [^\s]
> \s	A whitespace character: [ \t\n\x0B\f\r]
> X*	X, zero or more times
> .	Any character (may or may not match line terminators)
>
> Any character includes whitespace?  Why does my modified
> expression not replace every other "b"?

The problem here is, that the "b"s of input string are *also*
matched by the ".*", so each single match is likely a bit longer
than you expect it to be. I'm pretty sure that it still *did*
replace the very last "b".