CODEWITHSUNDEEP
dartdart-async
Fox32
Fox32

Reputation: 13560

Event with return value

I'm using the dart:async library to do some data processing. I'm adding objects to a StreamController and another module is listening to the stream. Now I want, that the other module is returning a result of the processing back to the add call (as a future).

Here is some example code that should illustrate what I want to do (It doesn't work, because the add method doesn't return a future): 

final controller = new StreamController();

controller.stream.listen((a) {
  // Do something with a, after that return something:
  return 42;
});

final aFuture = controller.add(new A());

aFuture.then((result) {
  // result == 42
});

Is something like this possible with dart:async, another library, or do I need to write my own classes?

PS: An alternative would be the following, but it would be 'more complicated' to use that a simple return:

final controller = new StreamController();

controller.stream.listen((container) {
  // Do something with container.a, after that return something:

  container.completer.complete(42);
});


final completer = new Completer();
controller.add(new Container(new A(), completer));

completer.future.then((result) {
  // result == 42
});

Upvotes: 0

Views: 292

Answers (1)

Allan
Allan

Reputation: 83

You could use a wrapper around the StreamController like:

class StreamControllerWrapper{
  MessageBox mb;
  StreamController controller;
  StreamControllerWrapper(this.controller){
    mb = new MessageBox();
    controller.stream.listen((a) {
      var replyTo = a['replyTo'];
      // Do something with a, after that return something:
      replyTo.add(42);
    });
  }
  Future add(msg){
    Completer c = new Completer();
    mb.stream.listen((reply){
      c.complete(reply);
    });
    controller.add({'content':msg, 'replyTo':mb.sink});
    return c.future;
  }
}

And then invoke the behavior like this:

final controllerWrapper = new StreamControllerWrapper(new StreamController());
controllerWrapper.add(new A())
  .then((result){
    print(result);
  });

Upvotes: 1

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