Damian Conway <[EMAIL PROTECTED]> writes: > Colin exemplifies: > > > $a = 1; > > @a = (1); > > @b = (1, 2, 3); > > @c = (4, 5, 6); > > > > $a = $a ^+ @b; > > @a = @a ^+ @b; > > > > print $a; # 7 > > No. It will (probably) print: 4. Because: > > $a = $a ^+ @b; > > becomes: > > $a = ($a,$a,$a) ^+ @b; > > which is: > > $a = (1,1,1) ^+ (1,2,3); > > becomes: > > $a = (2,3,4); > > which is: > > $a = 4;
Hmm... I thought that $a = $a ^+ @b becomes: $a = [$a, $a, $a] ^+ [1, 2, 3] $a = [1, 1, 1] ^+ [1, 2, 3] $a = [2, 3, 4] Or am I mistaken about the new perl6 syntax for $a = @b; Or does the hyper operator 'listify' the stuff it's working on? -- Piers
