type 'Future<List<Appointment>>' is not a subtype of type 'List<Appointment>' in type cast - flutter

Question

The error should be clear but I'm unsure how to go around it.
Basically I have a Stream builder I'm calling every second by getData() method to update my SfCalendar with new data.
Stream<DataSource> getData() async* {
await Future.delayed(const Duration(seconds: 1)); //Mock delay
List<Appointment> appointments = foo() as List<Appointment>;
List<CalendarResource> resources = bar() as List<CalendarResource>;
DataSource data = DataSource(appointments, resources);
print("Fetched Data");
yield data;
}
But my appointments method foo() is of type Future<List> and not List.
Future<List<Appointment>> foo() async {
var url0 = Uri.https(
"uri",
"/profiles.json");
List<Appointment> appointments = [];
try {
final response = await dio.get(url0.toString());
//final Random random = Random();
//_colorCollection[random.nextInt(9)];
response.data.forEach((key, value) {
appointments.add(
Appointment(
id: int.parse(
value["id"],
),
startTime: DateTime.parse(value["startTime"]),
endTime: DateTime.parse(value["endTime"]),
),
);
});
} catch (error) {
print(error);
}
return appointments;
}
That is what the error should be telling, yes?
I tried removing the Future cast from foo() appointments but then I can't use async.
I also tried returning Future.value(appointments) but same error.
This is where I call my Stream in initState():
#override
void initState() {
super.initState();
print("Creating a sample stream...");
Stream<DataSource> stream = getData();
print("Created the stream");
stream.listen((data) {
print("DataReceived");
}, onDone: () {
print("Task Done");
}, onError: (error) {
print(error);
});
print("code controller is here");
}
Thank you, please help when possible

Answer 1

Just like JavaScript, async functions always return a Future. That's why you can't use async when you remove Future from the return type.
Since you're not waiting for that Future to resolve, you're actually trying to cast a Future to a List, which isn't a valid cast. All you should need to do is wait for the function to finish so it resolves to a List:
List<Appointment> appointments = await foo() as List<Appointment>;
and, since your return type is Future<List<Appointment>>, you don't actually need to cast the result.
List<Appointment> appointments = await foo();