Given:
> $a = 1; > @b = (1, 2, 3); Damian suggested that: > $a = $a ^+ @b > > becomes: > > $a = ($a, $a, $a) ^+ (1, 2, 3) > $a = (1, 1, 1) ^+ (1, 2, 3) > $a = (2, 3, 4) > $a = 4; Whereas Piers thought that: > $a = $a ^+ @b > > becomes: > > $a = [$a, $a, $a] ^+ [1, 2, 3] > $a = [1, 1, 1] ^+ [1, 2, 3] > $a = [2, 3, 4] On this matter I can categorically state...that it is either the former, or the latter, or something else entirely. ;-) > Or am I mistaken about the new perl6 syntax for > > $a = @b; No. That *is* the same as C<$a = \@b>. > Or does the hyper operator 'listify' the stuff it's working on? That is indeed the question. Whether hyperoperators hyperoperate on lists or on arrays, and which of those they build if build they must. And it's a question that Larry alone can rule upon. So let me just find three small bits of wood and 4cc of mouse blood and we'll jolly well ask him... Damian
