On Tue, Nov 22, 2005 at 01:09:40AM +0100, Ruud H.G. van Tol wrote:
> >>>> 's/$/foo/' becomes 's/<after .*>/foo/'
> >>>
> >>> Uh, no, because <after> is still a zero width assertion.  :-)
> >>
> >> That's why I chose it. It is not at the end-of-string?
> >
> > Because ".*" matches "", /<after .*>/ would be true at
> > every position in the string, including the beginning,
> > and this is where "foo" would be substituted.
> I expected greediness, also because <after .*?> could behave non-greedy.
> ...
> But why does <after .*> behave non-greedy?

I think you may be misreading what <after .*> does -- it's a lookbehind
assertion.  An assertion such as <after pattern> attempts to match
pattern to the sequence immediately preceding the current match position.
It does not mean "skip over pattern and then match whatever comes

The greediness of the .* subpattern in <after .*> doesn't affect
things at all -- <after .*> is still a zero-width assertion.
Since ".*" can match at every position, <after .*> will be
a successful zero-width match (i.e., a null string) at every
position in the target string, including the beginning.

So, s/<after .*>/foo/  matches the first null string it finds 
-- the one at the beginning of the string -- and replaces it 
with "foo".  It's the same as if you had written s/<null>/foo/,
since <after .*> and <null> will both end up matching exactly
the same (i.e., a zero-width string at any position).

If this still doesn't make any sense, contact me off-list and
I'll try and explain it there.


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