Catch error using catchError in a future and throw another type

Question

I am not sure if this type of error handling and abstraction is done in a wrong way.

Future<void> _refresh() {
  return Future(() => throw someError)
       .catchError((error) {
         // maybe do something here
         throw abstractedError; //or even the same error
      });

Being able in another place to use it accordingly:

// in a widget/another place

void doSomething() { 
   _refresh()
     .then((_) => bla())
     .catchError((error) {
      //showSomeAlert() or handleSomething()
  });
}

Answer 1

We have in the flutter framework the description for the catch function:

• Handles errors emitted by this [Future].
• • This is the asynchronous equivalent of a "catch" block. ... Future catchError(Function onError, {bool test(Object error)});

---

When chaining futures I would advise against of using:

throw error;

and instead use:

return Future.error(SecondError());

This is because if you chain a future and expect to catch the error using the catchError future you will have the following problem.

Future<void> _refresh() {
  return throw Exception("Exception");
}

void main() {
    _refresh() // .then((_) => print("bla"))
    .catchError(() => print("not able to be reached"));
}

And you will get an error Uncaught exception: Exception: Exception.

This is similar when using RX (in general) and instead of throwing you send down the chain a Sigle.error (or any other observable with error).

---

TLDR:

When working with Futures and chaining them (including catchError) use Future.error(Exception("Exception")) to handle errors.

If you are using throw make sure or surround with try catch or Future(() -> throw ... )