Simplest way to convert an optional String to an optional Int in Swift

Question

It seems to me there ought to be a simple way to do an optional conversion from a String to an Int in Swift but I can't figure it out.

value is a String? and I need to return an Int?.

Basically I want to do this, but without the boilerplate:

return value != nil ? Int(value) : nil

I tried this, which seems to fit with Swift's conventions and would be nice and concise, but it doesn't recognize the syntax:

return Int?(value)

Answer 1

You can use the nil coalescing operator ?? to unwrap the String? and use a default value "" that you know will produce nil:

return Int(value ?? "")

---

Another approach: Int initializer that takes String?

From the comments:

It's very odd to me that the initializer would not accept an optional and would just return nil if any nil were passed in.

You can create your own initializer for Int that does just that:

extension Int {
    init?(_ value: String?) {
        guard let value = value else { return nil }
        self.init(value)
    }
}

and now you can just do:

var value: String?
return Int(value)

Answer 2

You can use the flatMap() method of Optional:

func foo(_ value: String?) -> Int? {
    return value.flatMap { Int($0) }
}

If value == nil then flatMap returns nil. Otherwise it evaluates Int($0) where $0 is the unwrapped value, and returns the result (which can be nil if the conversion fails):

print(foo(nil) as Any)      // nil
print(foo("123") as Any)    // Optional(123)
print(foo("xyz") as Any)    // nil