相关疑难解决方法(0)

在m4更新之后，StreamController.broadcast()被删除，我发现asBroadcastStream不能被多次调用。这意味着您必须存储流（如 Ugly 类中那样），而不是仅使用 getter（如 Pretty 类中那样）。

这是设计使然，还是我做错了？

void main() {
  Pretty p = new Pretty();

  //This does not work.
  p.onChange.listen((n) => print(n));
  //This second call throws: "Bad state: Stream already has subscriber". Why?
  p.onChange.listen((n) => print(n));

  //This does work.
  Ugly u = new Ugly();
  u.onChange.listen((n) => print(n));
  u.onChange.listen((n) => print(n));
}

class Pretty{
  StreamController<int> _streamCtrl = new StreamController<int>();
  Stream<int> get onChange => _streamCtrl.stream.asBroadcastStream();
}

class Ugly{
  StreamController<int> _streamCtrl = new StreamController<int>();
  Stream<int> _onChange;
  Stream<int> get onChange => _onChange;

  Ugly(){
    _onChange …