About Future generic type

Books Dart Apprentice: Beyond the Basics
1

I have a question about the Future generic type below
The code in the book is:

print('Before the future');

final myFuture = Future<int>.delayed(
  Duration(seconds: 1),
  () => 42,
)
    .then(
      (value) => print('Value: $value'),
    )
    .catchError(
      (Object error) => print('Error: $error'),
    )
    .whenComplete(
      () => print('Future is complete'),
    );

print('After the future');

myFuture is an instance of Future class. The type is inferred as Future<void>. But the right hand side is written as Future<int>.delayed( … How can this be possible? Or how do I understand the code above?

1 Like
2

The Future<int>.delayed constructor returns a Future<int>. The int refers to the value returned by () => 42. Since Dart can infer this type you can use Future.delayed directly without specifying the type. It’s still inferred to be Future<int>.

The confusion perhaps comes from the methods that are called on that returned future. Let me rewrite the code like so:

final myFuture = Future<int>.delayed(
  Duration(seconds: 1),
  () => 42,
);
final thenFuture = myFuture.then(
  (value) => print('Value: $value'),
);
final catchErrorFuture = thenFuture.catchError(
  (Object error) => print('Error: $error'),
);
final whenCompleteFuture = catchErrorFuture.whenComplete(
  () => print('Future is complete'),
);

The then method returns a Future<void>. So do the catchError and whenComplete methods. That’s why you see the inferred Future<void> for the original myFuture. The myFuture in my refactored code above has the expected Future<int> type.

1 Like
3

Thanks @suragch , your re-written code clears my thought.

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