'The return type 'void' isn't assignable to 'FutureOr<T>', as required by 'Future.catchError'
I have some code like this:
File("foo.txt").readAsString().catchError((e)=>print(e));
The compiler is complaining
info: The return type 'void' isn't assignable to 'FutureOr<T>', as required by 'Future.catchError'.
I can't seem to give it what it wants and can't find a single clear usage example in any of the docs (just a long issue in git about how many ways there are to mis-use this). If I take the docs at face value, I should be able to return a bool, or a future, neither make the analyzer happy.
How do I provide this FutureOr?
Solution 1:[1]
The documentation for Future.catchError could be a lot clearer, but the relevant part is:
onErroris called with the error and possibly stack trace, and the returned future is completed with the result of this call in exactly the same way as forthen'sonError.
Cross-referencing to the documentation for Future.then, the relevant portion is:
The
onErrorcallback must return a value or future that can be used to complete the returned future, so it must be something assignable toFutureOr<R>.
Since File.readAsString returns a Future<String>, your catchError callback also must return a Future<String>. Examples of doing that:
File("foo.txt").readAsString().catchError((e) {
print(e);
return Future.value('');
});
File("foo.txt").readAsString().catchError((e) async {
print(e);
return '';
});
Logically, this makes sense; because given:
String value = await File("foo.txt").readAsString().catchError(...);
then if readAsString succeeds, value should be assigned a String. If it fails, since you catch the exception without rethrowing it, value still needs to be assigned a String.
Put another way, your code is equivalent to:
Future<String> readFoo() async {
try {
return await File("foo.txt").readAsString();
} catch (e) {
print(e);
}
// Oops, missing return value.
}
In general, I strongly recommend using async/await with try-catch instead of using .catchError, which would avoid this confusion.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 |
