Richard Proctor's RFC166 says:
> =head2 Matching Not a pattern
>
> (?^pattern) matches anything that does not match the pattern. On
> its own, one can use !~ etc to negatively match patterns, but to
> match a pattern that has foo(anything but not baz)bar is currently
> difficult. With this syntax it would simply be /foo(?^baz)bar/.
The problem with this proposal is that it's really unclear what it
means.
The reason we don't have this feature today is not that it has never
been thought of before. People have thought of this a hundred times.
The problem is that nobody has ever figured out how it should work.
I don't mean that the implemenation is difficult. I mean that nobody
understand what such a a feature actually means. Richard doesn't say
this in his RFC, even for the simple examples he raises. He just
assumes that it will be obvious, but it isn't.
"foo-bazbar" =~ /foo(?^baz)bar/ # true or false?
"foo-baz-bar" =~ /foo(?^baz)bar/ # true or false?
OK, I'm going to try to invent a meaning for (?^baz). I'm going to
choose what appears to be a reasonable choice, and see what happens.
Let's suppose that what (?^baz) means is "match any substring that is
not 'baz'." That is a reasonably clear meaning. Then it behaves like
(.*)(?{$1 ne 'baz'}) does today. Then the examples above are both
true.
Now let's see how that choice works out.
"foobaz" =~ /foo.*(?^baz)/
This is TRUE, because "foo" matches "foo", ".*" matches "baz", and
"(?^baz)" matches the empty string at the end, which is a substring
that is not "baz".
In fact, with this apparently reasonable choice of meaning for
(?^baz), /foo.*(?^baz)/ will match anything that /foo.*/ will. The
(?^baz) has hardly any effect at all.
It is a good thing that we did not implement it that way, because it
is sure to become an instant FAQ: "Why does /foo.*(?^baz)/ match
'foobaz'?" You are going to see this question in comp.lang.perl.misc
every week.
So this choice I made for the meaning of (?^baz) appears to have been
the wrong one. I could go on and make a different reasonable-seeming
choice and show what was wrong with it, but I don't want to belabor my
point, which is:
Every choice anyone has ever made for the meaning of (?^baz) has
always been the wrong one for one reason or another. So without a
detailed explanation of what (?^baz) might mean, suggesting that Perl
6 have one is not helpful.
