In Swift, how do I convert String to Int for base 2, although base 36 is like Long.parseLong in Java?

Here parseLong()

at Swift. Note that the function returns Int?

(optional Int

) which must be unwrapped for use.

// Function to convert a String to an Int?.  It returns nil
// if the string contains characters that are not valid digits
// in the base or if the number is too big to fit in an Int.

func parseLong(string: String, base: Int) -> Int? {
    let digits = Array("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ")

    var number = 0
    for char in string.uppercaseString {
        if let digit = find(digits, char) {
            if digit < base {
                // Compute the value of the number so far
                // allowing for overflow
                let newnumber = number &* base &+ digit

                // Check for overflow and return nil if
                // it did overflow
                if newnumber < number {
                    return nil
                }
                number = newnumber
            } else {
                // Invalid digit for the base
                return nil
            }
        } else {
            // Invalid character not in digits
            return nil
        }
    }
    return number
}

if let result = parseLong("1110", 2) {
    println("1110 in base 2 is \(result)")  // "1110 in base 2 is 14"
}

if let result = parseLong("ZZ", 36) {
    println("ZZ in base 36 is \(result)")   // "ZZ in base 36 is 1295"
}