En r�ponse � "Keith C. Ivey" <[EMAIL PROTECTED]>:
> Marcelo E. Magallon <[EMAIL PROTECTED]> wrote:
>
> > A question of my own: why doesn't
> >
> > s/\B.\B/$&$&/g
> >
> > work as I expect, namely abcd -> abbccd. I really can't figure it
> > out by reading the docs.
>
> I wondered that too. I figured it was because \B wouldn't match
> twice at the same place, but then I saw that s/\B./$&$&/g didn't work
> as expected either.
What you want is:
s/(?!^).(?!$)/$&$&/g
Now, can we shorten it?
s/\B.(?=\B)/$&$&/g
Anything better?
--
Philippe BRUHAT - BooK
When you run from your problem, you make it that much harder for good
fortune to catch you, as well. (Moral from Groo The Wanderer #14 (Epic))
