How to get the path in file of a different file of the module?

Question

How do I get the path of another file of the module?

I tried it this way but I don't know where to get the $here and I don't know if it would work if I had the $here?

use v6;
unit class Aaa::Bbb::Ccc;

my $here = '.../lib/Aaa/Bbb/Ccc.pm6'.IO;
my $bin = $here.parent.parent.parent.add( 'bin' ).add( 'app' );

use Pod::Load;
my $pod = load( $bin );
# use the $pod

When I try this, $pod is empty:

use Pod::Load;
my $pod = load( $*PROGRAM );
say $pod.perl;  # $[]

Answer 1

There is no such thing as a "another file of the module". A module would represent only a single file; a distribution represent multiple modules and thus multiple files. Additionally a module cannot assume it is being loaded from a file system -- it might be loaded into memory out of a tar file or socket, and if a module were to try to access e.g. "../MyOtherModule.pm6".IO it would obviously fail.

What you can safely do is get at the content of any module in the same distribution:

unit class Aaa::Bbb::Ccc;

my $bin-code = $?DISTRIBUTION.content("bin/app").open.slurp;

Answer 2

Update: As @ugexe pointed out, if you intend on installing your module this will not work.

---

$?DISTRIBUTION seems like a good solution.

Otherwise an alternative is to use the $?FILE variable. You would need to export this from your module if want to use it somewhere else in your program:

unit module MyModule;

our sub myDIR() {
    $?FILE.IO.dirname;
}

use MyModule;
myDIR()~"/relative/path/to/somefile