Swift Optional Patterns

Question

Swift optional pattern allows you to use case let like this:

var arrayOfOptional: [Int?] = [1, 2, nil, 4]

for case let number? in arrayOfOptional {
  print("\(number)")
}

What confuses me is the let number? syntax. In optional binding, the unwrapped version does not have ?, but in case let, it does. How do you interpret this construct for it to make sense for you that number is unwrapped?

Functionally, what's the difference between the two below:

if let x = someOptional {
  print(x)
}

versus

if case let x? = someOptional {
  print(x)
}

Answer 1

I just tested your first code, never used for pattern matching before but here is what I assume:

var arrayOfOptional: [Int?] = [1, 2, nil, 4]

for case let number in arrayOfOptional {
    print("\(number)")
}
// will return 3 optional ints and a nil

for case let number? in arrayOfOptional {
    print("\(number)")
}
// will return only any values that could be unwrapped

I assume that this is a pattern which unwraps any optional value under the hood and only proceed if it could be unwrapped and will.

if case let x? = someOptional {
    print(x)
}

case let is used for pattern matching like switch x { case let ... }. In your example it will also try to unwrap an optional value. If it's a nil it will fail