Question 177165 in Flutter Question posted in
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Flutter – The value of expression of type 'void' can't be used in then()

How to fix this kind of errors when using then?

void handleSubmitWrapper() {
  final future = handleSubmit();
  future.then(context.loaderOverlay.hide()); // error
}

Future<void> handleSubmit() async {
  ...
}

This expression has a type of ‘void’ so its value can’t be used. Try
checking to see if you’re using the correct API; there might be a
function or call that returns void you didn’t expect. Also check type
parameters and variables which might also be void.dart

Tags:

2

Answers


  1. 0

    You must provide an anonymous function in your case, which has type FutureOr Function(void).

     future.then((void _) => context.loaderOverlay.hide());
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  2. 0

    Since you tagged this dart:async, lets have a look at what you could do when you don’t use that then anachronism:

    Future<void> handleSubmitWrapper() async {
  await handleSubmit();

  if (!context.mounted) return;

  context.loaderOverlay.hide();
}

    Bonus: you are now able to await this method. Or continue it with a then if you must. Or not. Your choice.

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