`Foo::Bar::<$quux>` is basically short for `::Foo::Bar::('$quux')` (except
the former is looked up at compile time)
So the way to do this is:
my $bar = 'Bar';
::Foo::( $bar ~ '::$quux' )
or
::Foo::($bar)::('$quux')
Note that the leading `::` are necessary for dynamic lookups.
El lun., 1 oct. 2018 a las 13:47, Richard Hogaboom (<
richard.hogab...@gmail.com>) escribió:
> Hmm... the ($bar) in Foo::($bar)::<$quux>; is an interpolation, but the
> <$quux> is just another way of writing $Foo::($bar)::quux;, not an
> interpolation, no?
>
> Right. It kinda is, but it should
Hmm... the ($bar) in Foo::($bar)::<$quux>; is an interpolation, but the
<$quux> is just another way of writing $Foo::($bar)::quux;, not an
interpolation, no?
On 10/1/18 7:41 AM, JJ Merelo wrote:
Thanks. I'll fix that.
WRT the original post, it looks like it should work, but apparently
Thanks. I'll fix that.
WRT the original post, it looks like it should work, but apparently can't.
The error should be selfexplanatory. Either you interpolate using :: or <>,
but not both...
JJ
El lun., 1 oct. 2018 a las 13:38, Richard Hogaboom (<
richard.hogab...@gmail.com>) escribió:
> Not
Not exactly, but close. The following line is exactly from the doc. It
works. It it works, then the offending(next line) line should work as well.
my$bar='Bar';
say$Foo::($bar)::quux;
# compound identifiers with interpolations; OUTPUT: «42»
sayFoo::($bar)::<$quux>; # won't compile -
Is that taken verbatim from the docs? I'll create an issue if that's the
case.
JJ
This does not compile; I think it should:
use v6;
class Foo {
class Bar {
our $quux = 42;
}
}
say $Foo::Bar::quux; # works - 42
say Foo::Bar::<$quux>; # works - 42
my $bar = 'Bar';
say $Foo::($bar)::quux; # works - 42
# shouldn't this work too?
say Foo::($bar)::<$quux>; #
