This is a follow up to my previous email re: http://www.perlmonks.org/index.pl?node_id=181980.
Since I'm trying to create "Perlog", a Prolog-like implementation in Perl, I've run across an issue which could cause some problems. Specifically, are variadic (varying number of arguments) 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 syntax, I'm going to have problems auto-generating code. I'm *assuming* that anything approaching variadic predicates would be handled with lists, thereby 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 them 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 __________________________________________________ Do You Yahoo!? Yahoo! Autos - Get free new car price quotes http://autos.yahoo.com
