A value of type 'Null' can't be returned by the 'onError' handler

user_12620572 0
Viewed 3282

I can't return a null from catchError handler of dart Future. I can do it using try catch but I need to use then catchError.

Using try catch

 Future<bool?> test() async {
    try {
      return await someFuture();
    } catch (e) {
      return null;
    }
  }

// Works without error

But when using then catchError

  Future<bool?> test() {
    return someFuture().catchError((e) {
      return null;
    });
  }

// Error: A value of type 'Null' can't be returned by the 'onError' handler because it must be assignable to 'FutureOr<bool>'

How do I return null if I encounter some error using then and catchError?

dart flutter asynchronous
2 Answers

You should specify someFuture() signature, which probably returns Future<bool>

Future<bool> someFuture() async

Method catchError must return the same future type it is called on. You can overcome this by forwarding the value to then and convert it to Future<bool?>:

Future<bool?> test() {
    return someFuture()
 .then((value) => Future<bool?>.value(value))
 .catchError((e) {
    return null;
});

}

user_2059550

This example works where I have made someFuture to return bool?:

Future<bool?> someFuture() async {
  throw Exception('Error');
}

Future<bool?> test() {
  return someFuture().catchError((Object e) => null);
}

Future<void> main() async {
  print('Our value: ${await test()}'); // Our value: null
}

If you can't change the return type of the someFuture method we can also do this where we creates a new future based on another future but where we specify that our type is nullable:

Future<bool> someFuture() async {
  throw Exception('Error');
}

Future<bool?> test() {
  return Future<bool?>(someFuture).catchError((Object e) => null);
}

Future<void> main() async {
  print('Our value: ${await test()}'); // Our value: null
}
user_19535150
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