[EMAIL PROTECTED] writes: : 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. : : ;-)
You're right. :-) Actually, I believe Piers is righter. The comma operator is no longer the C comma operator in Perl 6; it always builds a list object. If you happen to want only the last one, subscript with [-1]. But most of the time, the C comma operator is used in void context. : > Or am I mistaken about the new perl6 syntax for : > : > $a = @b; : : No. That *is* the same as C<$a = \@b>. More generally, \ and [] are really only required in list context, or to make an explicit reference to a scalar. Any list in scalar context becomes a list object. (Which, if evaluated in a numeric context, returns the length.) A list in list context also returns a list object. The list context merely promises that the list will be treated as flattened. It doesn't necessarily flatten it immediately as Perl 5 does. : > 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. Hyperoperators hyperoperate on objects. Some objects are scalars, and some objects are of higher dimensionality. The ^ is essentially a DWIM marker on the operator, which would otherwise assume both objects to be scalar. Larry
