Step 1 - Future value

Future<int> delayedOne() {
  return Future.delayed(const Duration(milliseconds: 100), () => 1);
}

Future<void> main() async {
  final n = await delayedOne();
  print(n);
}

• await pauses the current async function until the future completes; it does not block other tasks on the isolate.

Step 2 - async function

Future<String> fetchLabel() async {
  await Future.delayed(const Duration(milliseconds: 50));
  return 'ready';
}

Future<void> main() async {
  print(await fetchLabel());
}

Step 3 - Errors propagate

Future<void> boom() async {
  await Future.delayed(const Duration(milliseconds: 10));
  throw StateError('no');
}

Future<void> main() async {
  try {
    await boom();
  } catch (e) {
    print(e);
  }
}

Good habit

• Return Future<void> from main when using top-level await (Dart allows async main).

Practice tasks

• Write Future<int> sumAfterDelays() that awaits two Future.value delays and returns the sum of two ints.
• Call an async function without await and observe that main can finish before it runs; fix by awaiting or returning the future from main.