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Since binary "x" works like stated in perlop
Binary "x" is the repetition operator. In scalar context or if the left operand is not enclosed in parentheses, it returns a string consisting of the left operand repeated the number of times specified by the right operand. In list context, if the left operand is enclosed in parentheses or is a list formed by "qw/STRING/", it repeats the list. If the right operand is zero or negative, it returns an empty string or an empty list, depending on the context.
this operator resolves basically to !!. I'd call that one:

The Highlander Operator.

It resolves any gazillion operations on the rhs to just 1 (One).

()x!! then is the list asserter. Of course, a list cannot be asserted better than through the Highlander Himself - so it's full name is "The Highlander List Asserter" :-)

--shmem

_($_=" "x(1<<5)."?\n".q/)Oo.  G\        /
                              /\_/(q    /
----------------------------  \__(m.====.(_("always off the crowd"))."
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}

In reply to Re^2: Secret Perl Operators: The Highlander List Asserter by shmem
in thread Secret Perl Operators: the boolean list squash operator, x!! by Aristotle

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