This is a follow up to my previous email re:  http://www.perlmonks.org/index.pl?node_id=1819
80.

Since I'm trying to create "Perlog", a Prolog-like implementation in Perl, I've run across a
n
issue which could cause some problems.  Specifically, are variadic (varying number of argume
nts)
predicates allowed in Prolog?  In other words, can someone do this:

  foo( bar, baz ).
  foo( bar, quux, camel ).

I think I have the data structure issue figured out, but if I have to allow for the above sy
ntax,
I'm going to have problems auto-generating code.

I'm *assuming* that anything approaching variadic predicates would be handled with lists, th
ereby
avoiding the problem:

  foo( bar, [ baz ] ).
  foo( bar, [ quux, camel ] ).

If variadic predicates are allowed, what would they be used for?  If I can safely exclude th
em
from the Perl implementation, this would make life much easier.

Cheers,
Curtis "Ovid" Poe

=====
"Ovid" on http://www.perlmonks.org/
Someone asked me how to count to 10 in Perl:
push@A,$_ for reverse q.e...q.n.;for(@A){$_=unpack(q|c|,$_);@a=split//;
shift@a;shift@a if $a[$[]eq$[;$_=join q||,@a};print $_,$/for reverse @A

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