Turns out I misspoke anyway; << >> isn't enough by itself.
$ p6 'my @x=<a b c>; my @y=<<1 2 @x 3 4>>; say @y; dd @y;'
@y will contain (1, 2, '@x', 3, 4). (Actually, the numbers will be things
that are simultaneously numbers and strings, showing up as IntStr.new(1,
'1') and such.) '@x' is a two-character string, not the list @x or its
contents.
pyanfar Z$ 6 'my @x = <a b c>; my @y = <<1 2 @x[] 3 4>>; dd @y'
Array @y = [IntStr.new(1, "1"), IntStr.new(2, "2"), "a", "b", "c",
IntStr.new(3, "3"), IntStr.new(4, "4")]
Which flattens because it's interpolating values, since quoting constructs
produce string-like things or lists of string-like things (role Stringy),
not lists of lists. If you want a list of lists, you are better off just
using lists directly, not the quoting constructs that are intended to give
you lists of strings. This is similar to Perl 5, where qw() also doesn't
expand variables:
pyanfar Z$ perl -MData::Dumper -E 'my @x = qw(a b c); my @y = qw(1 @x
2); say Dumper(\@y)'
$VAR1 = [
'1',
'@x',
'2'
];
On Fri, Oct 5, 2018 at 11:27 PM ToddAndMargo via perl6-users <
perl6-users@perl.org> wrote:
> On 10/5/18 6:28 PM, Brandon Allbery wrote:
> > You can't with that, because < ... > acts like single quotes; '@x' there
> > is a string literal, not the list @x. If you use << ... >> then it act
> > like double quotes, and will use the list instead.
>
> I am sorry. With the top posting, I can't tell
> what you are talking about. I also can't tell which
> of the three question I asked that you are
> answering
>
> :'(
>
> >
> > On Fri, Oct 5, 2018 at 9:23 PM ToddAndMargo via perl6-users
> > <perl6-users@perl.org <mailto:perl6-users@perl.org>> wrote:
> >
> > On 10/5/18 6:09 PM, Brandon Allbery wrote:
> > > That's where the | comes in. If you say
> > >
> > > my @b = (1, |@a, 2);
> > >
> > > then you get (1, 'a', 'b', 'c', 2) like in Perl 5. But you can
> > specify
> > > what gets flattened, so you can choose to flatten some arrays but
> > not
> > > others if you need to for some reason:
> > >
> > > my @b = (1, |@b, 2, @b, 3);
> > >
> > > gives you (1, 'a', 'b', 'c', 2, ('a', 'b', 'c'), 3). This often
> > matters
> > > in parameter lists, if you want one of the parameters to be the
> list
> > > itself instead of it being spread across multiple parameters. In
> > Perl 5
> > > you had to say \@b to pass a ref to the list instead, and the sub
> > had to
> > > know it was getting a scalar containing an arrayref and needed to
> > > dereference it (if you don't know, don't ask; it's ugly).
> >
> > Am I correct in my assumption?
> >
> > A little off the question, but how do I address something
> > inside a nested array?
> >
> > In the following, how do I get at the "b"
> >
> >
> > $ p6 'my @x=<a b c>; my @y=<1 2 @x 3 4>; say @y; dd @y;'
> >
> > [1 2 @x 3 4]
> >
> > Array @y = [IntStr.new(1, "1"), IntStr.new(2, "2"), "\@x",
> > IntStr.new(3,
> > "3"), IntStr.new(4, "4")]
> >
> > Also, how do I flatten @y above?
> >
> > Thank you for the help!
> > -T
> >
> >
> >
> > --
> > brandon s allbery kf8nh
> > allber...@gmail.com <mailto:allber...@gmail.com>
>
>
> --
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> Computers are like air conditioners.
> They malfunction when you open windows
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>
--
brandon s allbery kf8nh
allber...@gmail.com
