swift - sort an array of objects by their optional boolean property without force unwrapping

Question

I can sort this array of store objects by their 'flagship' boolean property, but how can I safely unwrap the 'flagship' property first?

let flagshipStores = self.stores.sort {
    $0.flagship! && !$1.flagship!
}

Answer 1

let flagshipStores = self.stores.sort {
    guard let flagship0 = $0.flagship, let flagship1 = $1.flagship else { return false }
    return flagship0 && !flagship1
}

Answer 2

One more approach: turn the Bool? into an Int, then compare the Ints. You get to specify how a nil value compares to non-nil values.

For instance, this sorts nil values before both false and true:

stores.sort { Int($0.flagship ?? -1) < Int($1.flagship ?? -1) }

This sorts nil values after both false and true:

stores.sort { Int($0.flagship ?? 2) < Int($1.flagship ?? 2) }

You can use the same pattern to make nil compare the same as true or the same as false. It's up to you.

Answer 3

Here's another approach.

You can use flatMap which will remove nil objects and unwrap those that are present. Then, the force unwrap will be safe to sort:

let flagshipStores = stores.flatMap({ return $0.flagship ? $0 : nil }).sort {
    $0.flagship! && !$1.flagship!
}

This will remove stores with a nil flagship from the array.

Answer 4

How about:

$0.flagship == true && $1.flagship != true

The left side will succeed if the value is not nil and is true, the right side will succeed if the value is either nil or false.