Function throws AND returns optional.. possible to conditionally unwrap in one line?

Question

I am using an SQLite library in which queries return optional values as well as can throw errors. I would like to conditionally unwrap the value, or receive nil if it return

Answer 1

I like Martin's answer but wanted to show another option:

if let value = (try? getSomething()) ?? nil {

}

This has the advantage of working outside of if, guard, or switch statements. The type specifier Any? isn't necessary but just included to show that it returns an optional:

let value: Any? = (try? getSomething()) ?? nil

Answer 2

Update: As of Swift 5, try? applied to an optional expression does not add another level of optionality, so that a “simple” optional binding is sufficient. It succeeds if the function did not throw an error and did not return nil. val is then bound to the unwrapped result:

if let val = try? getSomething() {
    // ...
}

---

(Previous answer for Swift ≤ 4:) If a function throws and returns an optional

func getSomething() throws -> Value? { ... }

then try? getSomething() returns a "double optional" of the type Value?? and you have to unwrap twice:

if let optval = try? getSomething(), let val = optval {

}

Here the first binding let optval = ... succeeds if the function did not throw, and the second binding let val = optval succeeds if the return value is not nil.

This can be shortened with case let pattern matching to

if case let val?? = try? getSomething() {

}

where val?? is a shortcut for .some(.some(val)).