Dart: How to return a known, constant, or literal value in a Future

Sometimes a Dart or Flutter API may require you to return a Future even if you already have a known, constant, or literal value. If/when that happens, you can use this approach:

return new Future(() { return 42; });

In this example the known value is 42, but it could have been a string like "Hello", a boolean like true, or any other known or literal value.

I was reminded of this when I saw the following Dart Future/delayed code, and asked why it was written the way it was:

return Future.delayed(Duration(milliseconds: 0))
    .then((onValue) => 42);