> On 03 Jan 2016, at 10:44, Jules Field <ju...@jules.uk> wrote: > The snag is this: > > my @a = (1, (2,3), 4); > [1 (2 3) 4] > > I asked for a List, but got given an Array. So I can do this:
If like that, you really have asked for an Array (@a) to be assigned with the values of a List (1, (2,3), 4). If you want @a to be a List, you need to *bind* it to the List. $ 6 'my @a = (1, (2,3), 4); say @a.WHAT' (Array) $ 6 'my @a := (1, (2,3), 4); say @a.WHAT' (List) > > (1, (2,3), 4).flat; > (1 2 3 4) > > but not this: > > @a.flat; > (1 (2 3) 4) .flat returns a Seq $ 6 'my @a := (1, (2,3), 4); say @a.flat.WHAT' (Seq) > And now @a.flat is a list. But if I assign the result to @b then this happens: > > my @b = @a.flat; > [1 (2 3) 4] > > Now it's turned into an Array again! Again, you need to bind if you want @b to be a List. Because you can’t bind a Seq, it needs to be coerced to a List: $ 6 'my @a = (1, (2,3), 4); my @b := @a.flat.list; say @b.WHAT' (List) > How can it be right that if I split 1 line of code into 2 by creating an > intermediary variable, the code behaves totally differently? > If it's intentional, then what am I doing wrong in my trivial example above? Because the @ sigil forces an Array by default. You don’t have to use an @ sigil though: $ 6 'my $a = (1, (2,3), 4); my $b = $a.flat.list; say $a.WHAT; say $b.WHAT' (List) (List) > Many thanks for your help! Hope this helps… Liz
