On 2019-06-20 16:15, ph10 wrote:
> You can see all this by making use of the "auto-callout" feature
Thanks a lot, Philip. I quite well understand what is really happened.
My concern is about how this is documented.
>In the first example, the same thing happens, but after (?=b) ismatched,
> \z fails so there is a backtrack into the last iteration of theloop.
Sorry I only know a basic english and don't know many nuances of this
language.
When "loop is forcibly broken" I assume that there is no more loop. How
can we return (backtrack) to it? Is there possibility to return inside a
cycle after operator "brake" in any programming language?
May be another words can be find?
Or I don't quite understand some nuances of english "broken"?
Second of my little concern is that "X*\z" and "X*" both matches and
matches are different.
I understand why it is from procedural point of view. Know that is
Perl-compatible. But from logical point of view is quite unnaturally.
