invalid_return_type_for_then

A value of type '{0}' can't be returned by the 'onError' handler because it must be assignable to '{1}', as required by 'Future.then'.

The return type '{0}' isn't assignable to '{1}', as required by 'Future.then'.

The analyzer produces this diagnostic when the return type of the onError argument in Future.then is incompatible with the return type of the onValue argument. At runtime, Future.then attempts to return the value from the onError handler as the future's result, which throws another exception.

The following code produces this diagnostic because the onError handler returns a String instead of an int:

void f(Future<int> future) {
  future.then((_) => 0, onError: (e, st) => 'c');
}

The following code produces this diagnostic because the return type of the onError argument (cb) is incompatible with the return type of onValue:

void f(Future<int> future, String Function(dynamic, StackTrace) cb) {
  future.then<int>((_) => 1, onError: cb);
}

If the onValue handler returns the correct type, then change the onError handler to match:

void f(Future<int> future) {
  future.then((_) => 0, onError: (e, st) => -1);
}

If the onError handler is correct, then change the onValue handler to match:

void f(Future<String> future) {
  future.then((_) => 'a', onError: (e, st) => 'c');
}

If both handlers are correct, then change the future type to match:

void f(Future<String> future) {
  future.then<Object>((_) => 0, onError: (e, st) => 'c');
}