how to handle Result<nil, nil> ? about result HOT 8 CLOSED

Have you tried Result<Void, MyErrorType>?

enum MyErrorType: ErrorType {
    case None
}

I was wondering kind of the same some weeks ago @ http://stackoverflow.com/questions/30875889/implement-an-equatable-void-none-type/33142599#33142599 :-)

from result.

An empty array is perfectly reasonable for the result if your query returns a collection of values. If it's a scalar that's optional, define it as such, e.g. Result<Int?, SomeError>

from result.

tried that, but here is the weird result:

enum Whatever: ErrorType {
    case e1, e2
}

func test1() -> Result<String?, Whatever> {
    return .Success(nil)
}

if let s = test1().value {
    print("s should not be nil: \(s)")
} else {
    print("s is nil")
}

and we got:

s should not be nil: nil

from result.

That's because value returns T? which means String?? if you plug in your types. The if-let unwraps the outer optional of the returned Some(nil), hence printing nil.

from result.

Sure, but the question remain that how do we elegantly retrieve the actual value without verbose workaround

if let s1 = test1().value, s = s1 ...

if let s = (test1().value ?? nil) ...

Is there a better way to do this double-optional unwrapping? Can the value property auto unwrap the optional before returning? Ideally if let s = test1.value should work for either Result<String,Error> or Result<String?,Error>...

from result.

I wouldn't call it a "verbose workaround", that's just when dealing with multiple nested optionals. Alternatively, if you're using Swift 2 you could pattern match over the result and avoid value entirely. (This doesn't work as well in Swift 1.2 due to the Boxed values).

from result.

You can write the case as follows in Swift 2:

if case let s?? = test1().value {
    print("s should not be nil: \(s)")
} else {
    print("s is nil")
}

from result.

I'm going to go ahead and close this. If it's still an issue or if these solutions don't work for you, feel free to re-open.

from result.