I noticed that even a simple regex in a (??{...}) block doesn't work:

#!/usr/bin/perl
use strict;
use warnings;

$_ = "foo";

print "a" =~ m{.} ? "o\n" : "O\n";

if ( m/f (??{ "a" =~ m{.} ? 'o' : 'O' }) o/x){
    print "matched\n"
}
__END__
o
[download]

In your position I'd precompute the regex instead of trying to stuff it all in one line.

Hi Moritz,

Thx for the feedback.

Yes I've read the perlre. Unfortunately my client interpretes highly experimental by default as "Oh. So it's feasible". :-)

Precomputing the most inner block gives the same result.

With kind regards.

DirkDn

Unfortunately my client interpretes highly experimental by default as "Oh. So it's feasible". :-)

As a programmer it's your job to tell him otherwise.

If possible create an example that segfaults (for example try some fun with closures inside the block), and show that to your client ;-)

Precomputing the most inner block gives the same result.

Then you have to unroll one more level:

!/usr/bin/perl
use strict;
use warnings;

use Data::Dumper;
my @A = ( "ab", "abcd", "f" );
my %H = (
    'ab' => 0,
    'abcd' => 1,
#'def' => 2,
);

my $re_inner = join( '|', @A);
my $re_outer = join '|', grep  /$re_inner/gi, keys %H;
my @B = grep ( !/$re_outer/gi, @A );

print Dumper(\@B);
[download]

Disclaimer: I haven't actually understood what you try to achieve, maybe there's a better solution altogether.

From perlre for 5.10.0

Because Perl's regex engine is currently not re-entrant, interpolated code may not invoke the regex engine either directly with m// or s///), or indirectly with functions such as split.

Dear Monks Moritz, Ikegami & Hipowls,

A bit delayed reply since I was offline due to a flue.

Thanks a lot for this precious information. I've learned a lot from it.

With kind regards.
ddn123456

The syntax error you are getting is due to the presence of the regexp delimiter in the (??{...}) assertion. To make your original code with minimum change, change
@B = grep ( !/(??{join( '|', grep ( /(??{join( '|', @A)})/gi, keys %H ) })/gi, @A );
to
@B = grep ( !/(??{join( '|', grep ( m!(??{join( '|', @A)})!gi, keys %H ) })/gi, @A );

Update: hipowls notes that even if you fix the syntax error (as shown in this post), it won't work because the regexp engine is not re-entrant.