--marmot

Also see the OP for the "semi-predicate problem" discussion.

If this seems too theoretical for you, consider the practical task to do of a run-length encoding of sparsely set arrays. (undef is a real value)

Good point about the initial value -- I forgot about that.

It can be specially handled decently, though, something like:

my @new = my $p = shift @orig;
push @new, grep { $p = $_ or 1 if $p ne $_ } @orig;
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Shoot! I totally get the semi-predicate problem, and you stated the problem clearly in the OP, but I thought I had a brainwave and rewrote my UPDATED code, which had actually accounted for the problem.

So anyway, I re-read the entire thread and I sort of get why you'd want, say, a $^PRE special variable. But it is so trivial, I don't see why you'd bother. You are simply dealing with a list in pairwise fashion, using a previous value to evaluate a current value. Eliminating adjacent dupes is trivial...

    @new = map { $p ne $_ ? $p = $_ : () } @orig;
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except for an initial undefined value, but only because $p is initially undefined. So you define it and it works. And it's easy to remember.

    $p = 'supercalifragilistic';
    @new = map  { $p ne $_ ? $p = $_ : () } @orig;
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You could define a uniq_adj function...

    use strict;
    sub uniq_adj {
        my $p = 'supercalifragilistic';
        return map  { $p ne $_ ? $p = $_ : () } @_;
    }

    my @orig = (undef, undef, qw(a a b b c 0 c d d u u 0 0 "0" "0" '0'
+ '0'));
    my @new = uniq_adj @orig;

    print "'", join ("' '", @new), "'\n";
    __END__
    Prints --> '' 'a' 'b' 'c' '0' 'c' 'd' 'u' '0' '"0"' ''0''
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I printed with single quotes to show where undef's are being printed. Note that 0, "0", and '0' are preserved. It's very generalizable. I predefined $p, but isn't that less work than having Perl do it through a built-in? I read through the entire thread again and it seems to fit the bill. Is this kind of what you're looking for?

Regardless, cheers!

--marmot

Your surely aware that replacing undef with a "magic" string is not a solution of the semi-predicate problem, because the magic string can also be an element of the processed list.

Cheers Rolf