What is the equivalent to `thing = this() || that` in Swift 3? - swift3

Question

In most languages I can write something like this:
this = this() || that()
If this() returns something false, like integer 0, that() will be evaluated. Pretty common idiom.
Swift 3 doesn't automatically cast Int to Bool so that idiom doesn't work. What is a succinct way to do it in Swift 3?

Answer 1

There's no exact equivalent in Swift, as:
No other types than Bool cannot be converted to Bool
You cannot write a function returning Bool or Int
(In most languages is a little bit exaggerated, you cannot write such thing in Java, C# or many other strongly-typed languages.)
The most similar thing in Swift is nil-coalescing operator -- ??.
Assume this() returns Int? (aka Optional<Int>), and that() returns Int (non-Optional):
func this() -> Int? {
//...
if someCondition() {
return intValue
} else {
return nil
}
}
func that() -> Int {
//...
return anotherIntValue
}
And use nil-coalescing operator like this:
let result = this() ?? that()
In this assignment, if this() returns non-nil value, then that() is not evaluated, and that non-nil value is assigned to result. When this() returns nil, that() is evaluated and the value is assigned to result.

Answer 2

The nil coalescing operator as covered by #OOPer:s answer is the fit for purpose idiom here, but as a side-note, you could implement functionality as the one your describe by overloading the || operator, e.g. for types that conform to some type constraint. E.g. using Integer as type constraint:
extension Bool {
init<T : Integer>(_ value: T) {
self.init(value != 0)
}
}
func ||<T: Integer>(lhs: T, rhs: #autoclosure () -> T) -> T {
return Bool(lhs) ? lhs : rhs()
}
var thisNum = 0
let thatNum = 12
thisNum = thisNum || thatNum
print(thisNum) // 12