Do all Perl 6 quoting constructs have term precedence?

Question

The < > has term precedence. Here's the example from the docs:

say <a b c>[1];

I figured the same precedence would apply to all of the quoting operators. This works:

my $string = '5+8i';
my $number = <<$string>>;
say $number;

This interpolates $string and creates allomorphes (in this case a ComplexStr):

(5+8i)

But, if I try to index it like the example from the docs, it doesn't compile:

my $string = '5+8i';
my $number = <<$string>>[0];
say $number;

I'm not quite sure what Perl 6 thinks is happening here. Perhaps it's thinking it's a hyperoperator:

===SORRY!=== Error while compiling ...
Cannot use variable $number in declaration to initialize itself
at /Users/brian/Desktop/scratch.pl:6
------>     say $⏏number;
    expecting any of:
        statement end
        statement modifier
        statement modifier loop
    term

I can skip the variable:

my $string = '5+8i';
say <<$string>>[0];

But that's a different error that can't find the closing quotes:

===SORRY!=== Error while compiling ...
Unable to parse expression in shell-quote words; couldn't find final '>>'
at /Users/brian/Desktop/scratch.pl:8
------> <BOL>⏏<EOL>
    expecting any of:
        statement end
        statement modifier
        statement modifier loop

Answer 1

Jonathan has the answer in response to RT #131695.

The >>[] is a postfix operator to index a list, so it tries to use it as such. This is intended behavior. Fair enough, although I think the parser is being a bit too clever for regular code monkeys here.

Answer 2

I think this warrants a rakudobug email. I think the parser gets confused trying to interpret it as a hyper (aka >>.method). The following workaround seems to corroborate this:

my $string = '5+8i';
my $number = <<$string >>[0];  # note space before >>
say $number;

To satisfy your OCD, you could probably also put a space before $string.

And yes, whitespace is not meaningless in Perl 6.