Euclidean algorithm pseudocode conversion to Swift?

Question

I have been working on a function for reducing fractions in Swift, and came across the Euclidean algorithm for finding the greatest common factor (http://en.wikipedia.org/wiki/Euclidean_algorithm)

I converted the pseudo code into swift, but yet I am confused how this is going to give me the greatest common factor if it is returning a which I thought was supposed to be the numerator of the fraction. Any help on this would be greatly appreciated. Thanks!

Pseudocode:

function gcd(a, b)
    while b ≠ 0
       t := b
       b := a mod b
       a := t
     return a

Swift:

var a = 2
var b = 4

func gcd(a: Int, b: Int) -> Int {
    var t = 0
    while b != 0 {
        t = b
        let b = a % b
        let a = t
    }
    return a
}

println("\(a)/\(b)")

Console output: 2/4

Answer 1

When you do this

let b = a % b

you are creating another readonly variable b, which has nothing to do with the variable b from the outside scope. You need to remove both lets inside the loop, and make parameters modifiable by declaring them with var, like this:

func gcd(var a: Int, var b: Int) -> Int {
    var t = 0
    while b != 0 {
        t = b
        b = a % b
        a = t
    }
    return a
}

You can call your function like this:

let a = 111
let b = 259
println("a=\(a), b=\(b), gcd=\(gcd(a,b))")

This prints a=111, b=259, gcd=37

Answer 2

Taking @dasblinkenlight's answer and getting rid of t by using tuples for parallel assignment yields:

Swift 2.1:

func gcd(var a: Int, var _ b: Int) -> Int {
    while b != 0 {
        (a, b) = (b, a % b)
    }
    return a
}

gcd(1001, 39)  // 13

---

var parameters are deprecated in Swift 2.2 and will be removed in Swift 3. So now it becomes necessary to declare a and b as var within the function:

func gcd(a: Int, _ b: Int) -> Int {
    var (a, b) = (a, b)
    while b != 0 {
        (a, b) = (b, a % b)
    }
    return a
}