CODEWITHSUNDEEP
flutterdartasync-awaitfuture
无夜之星辰
无夜之星辰

Reputation: 6148

How to return the function in a closure in dart?

eg:

Future<int> _test1() async {
  return Future.delayed(Duration(seconds: 1)).then((value) {
    return 1;
  });
}

void _onTapButton() async {
  await _test1().then((value) {
    print('a');
    if (value == 1) {
      print('return');
      return;
    }
  });

  print('b');
}

When I call _onTapButton, console print a return b not a return.

That is to say the return in _test1().then didn't return the function _onTapButton.

Is there any way I can return function _onTapButton in _test1().then?

Upvotes: 1

Views: 1094

Answers (1)

Peter Koltai
Peter Koltai

Reputation: 9734

In your code then part is a callback that will be executed upon completion of your future, and return is meant in the scope of this callback, not your _onTapButton function.

If you want to wait for the future, and if it results 1, print return and return from _onTapButton, you can try this:

void _onTapButton() async {
   final value = await _test1();
   print('a');
   if (value == 1) {
      print('return');
      return;
   }
   print('b');
}

Upvotes: 2

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