A value of type 'Future<bool>' can't be returned from the function because it has a return type of 'Future<void>'

user_12483095 0
Viewed 1254

I have:

Future<bool> foo() async => true;

This is allowed:

Future<void> bar()  {
  return foo(); // Works
}

but this is not:

Future<void> baz() async {
  return foo(); // Error
}

In both bar and baz I'm returning a Future<bool>, but why first works but second fails?

Note: This question isn't about HOW to make it work but rather WHY one works and the other doesn't.

dart-null-safety dart
3 Answers

Dart has special rules for void returning functions: You are not allowed to return a value because it's declared as not returning any value, so actually trying to return a value is probably a mistake.

The same happens without Future:

void qux() {
  return true; // Error, can't return value from void function
}

gives the error

A value of type 'bool' can't be returned from the function 'qux' because it has a return type of 'void'.

You are allowed to have return e; statements, but only if the type of e is void, Null or dynamic. (And you shouldn't even do that, those are just allowed in order to make existing code work.) (A => e body is always allowed, because people like to use it as a shorthand for just { e; }, mainly because the formatter keeps it on one line. I still recommend using { e; } as body of a void function.)

Generalizing that to async functions with void future return types, you are not allowed to return an actual value from Future<void> ... async function. So, the only things you are allowed to return are void, dynamic, Null, Future<void>, Future<dynamic>, and Future<Null>. A Future<bool> is neither of those.

What you should write instead is:

Future<void> baz() async {
  await foo();
}
user_21566210

As I have observed when I am running the same code

Future<bool> foo() async => true;

Future<void> bar() {
  return foo(); // Works
}

Future<void> baz() async {
  // return foo(); // Error
}

void main() {
  print("direct =====${foo()}");
  print("bar==========${bar()}");
  print("bazz========${baz()}");
}

The response is

direct =====Instance of '_Future<bool>'
bar==========Instance of '_Future<bool>'
bazz========Instance of '_Future<void>'

this signifies that when the async keyword is not present it takes the return type of the returning value.

But of the bazz function it gives a quick fix to make the function

Future<Future<bool>>

so when async is added the function return type is taken as the main type

user_161864480 
Future<bool> foo() async => true;

First Case

Future<void> bar()  {
  return foo(); // Works
}

For the code above, the return of Future<void> bar() can return foo(), because the bar() not return directly the value of foo(), so the bar() can deal with foo() without no problem.

But, in the second case:

Future<void> baz() async {
  return foo(); // Error
}

The code above is return directly the generic type of foo() because the function of baz() include async tag, so it will be trouble with the baz() as Future<void>, which is we know that foo() as Future<bool>

user_172995180
Related