`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. (Wher
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 ra
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
can't
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 exa
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 - but
Is that taken verbatim from the docs? I'll create an issue if that's the
case.
JJ
