Can you clone a Perl 6 Proc?

Question

I was playing with this in 2018.01:

my $proc = Proc.new: :out;
my $f = $proc.clone;
$f.spawn: 'ls';
put $f.out.slurp;

It says it can't do it. It's curious that the error message is about a routine I didn't use and a different class:

Cannot resolve caller stdout(Proc::Async: :bin); none of these signatures match:
    (Proc::Async:D $: :$bin!, *%_)
    (Proc::Async:D $: :$enc, :$translate-nl, *%_)
  in block <unit> at proc-out.p6 line 3

Answer 1

Everything inherits a default clone method from Mu, which does a shallow clone, but that doesn't mean that everything makes sense to clone. This especially goes for objects that might hold references to OS-level things, such as Proc or IO::Handle. As the person who designed Proc::Async, I can say for certain that making it do anything useful on clone was not a design consideration. I didn't design Proc, but I suspect the same applies.

As for the error, keep in mind that the Perl 6 standard library is implemented in Perl 6 (a lot like in Java and .Net, but less like Perl 5 where many things that are provided by default go directly to something written in C). In this particular case, Proc is implemented in terms of Proc::Async. Rakudo tries to trim stack traces somewhat to eliminate calls inside of the setting, which is usually a win for the language user, but in cases like this can be a little less helpful. Running Rakudo with the --ll-exception flag provides the full details, and thus makes clearer what is going on.