I have written some code and I want to provide it in a package, but I want to also expose it to package consumers as a worker. For this purpose I have created a wrapper class, that runs an isolate internally and using the send command and listeners communicate with the Isolate to provide the functionality.
The problem arises when I want to use this wrapper class from bin
or web
directory: the Uri
provided is interpolated from the directory of the running/main Isolate
instead of from the package root. For bin
it is packagename|bin/
and for web
it is packagename|web
.
I would like to export this class to the consumers so they can chose an easier approach than to construct their own Isolate
, but I am not sure how to specify the main file that will be used in spawnUri.
Is there a way to specify the file so it will always be resolved to the correct file regardless of where the main Isolate is run from.
Structure:
// Exports the next file so the class in it will be package visible
packageroot -> lib/package_exports_code_that_spawns_isolate.dart
// This file should contain URI that always resolve to the next file
packageroot -> lib/code_that_spawns_isolate.dart
// The main worker/Isolate file
packageroot -> lib/src/worker/worker.dart
Thanks.
To refer to a library in your package, you should use a package:
URI.
Something like:
var workerUri = Uri.parse("package:myPackage/src/worker/worker.dart");
var isolate = await Isolate.spawnUri(workerUri,...);
It's not perfect because it requires you to hard-wire your package name into the code, but I believe it's the best option currently available.
The Isolate.spawnUri
function doesn't (and can't) resolve a relative URI reference wrt. the source file that called it - nothing in the Dart libraries depends on where it's called from, that's simply too fragile - so a relative URI isn't going to work. The only absolute URI referencing your worker is a package:
URI, so that's what you have to use.
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