Hi, I would like to give a perspective a bit different to what yary provided.
Your example here can be golfed down to: my $a = [1,2,3]; my @v = $a; say @v; # [[1,2,3],] The reason for this behavior is $a being a scalar container. Correspondingly, when you assign it to @v the assignment op see a single element, not a positional. So, it behaves correspondingly. Apparently: my @v = [1,2,3]; say @v; Do what's expected. The reason is [1,2,3] on RHS is a positional. It's a non-containerized value. So, if we put one and one together we come to the following assignment: my @v = $a<>; say @v; # [1,2,3] Because by stripping of the container with <> operator we get a positional on RHS. Now, if we unroll it all back your case, the only thing we should remember is that arrays containerize any value assigned into them. I.e.: constant foo = 4; my @v; @v[0] = foo; say 4.VAR.^name; # Int say @v[0].VAR.^name; # Scalar Therefore, when you do @first = @both[0] – it is equivalent to @v = $a in the first example. We have a documentation section on this: https://docs.raku.org/language/list#Itemization Best regards, Vadim Belman > On Jan 19, 2021, at 1:18 PM, Brian Duggan <bdug...@matatu.org> wrote: > > Hi Folks, > > I ran into this situation today, which seems counterintuitive: > > my @one = 1,2,3; > my @two = 4,5,6; > my @both = @one,@two; > my @first = @both[0]; > say @one.raku; > say @first.raku; > > output: > > [1, 2, 3] > [[1, 2, 3],] > > I was expecting @first and @one to be the same. > I discovered that I could instead write either of these -- > > my (@first) = @both[0]; > my @first := @both[0]; > > or I could change the @both assignment to be > > my @both := @one, @two; > > ..but I wonder if there's an idiomatic approach -- or > way of thinking about this -- that makes this flow more > intuitive. > > thanks > Brian >
