Apply void function on array - swift

Question

I am learning swift right now and wanted to write a quicksort algorithm for [Int]
func swap(array:[Int], x:Int, y:Int)->[Int] { //swap elements in array
var result:[Int] = array
result[y] = array[x]
result[x] = array[y]
return result
}
func split(array:[Int], u:Int, o:Int , p:Int)->Int { //where the sorting happens
let pivot:Int = array[p]
swap(array: array, x: p, y: o)
var pn:Int = u
for j in u...o {
if(array[j] <= pivot) {
swap(array: array, x: pn, y: j)
pn = pn+1
}
}
swap(array: array, x: pn, y: o);
return pn;
}
func quickSortR(array:[Int],u:Int ,o:Int) { //recursive call
if(u < o){
let p:Int = o
let pn:Int = split(array: array, u: u, o: o, p: p)
quickSortR(array: array,u: u,o: pn-1)
quickSortR(array: array,u: pn+1,o: o)
}
}
func quickSort(array:[Int]) { //initial call
quickSortR(array: array, u: 0, o: array.count-1)
}
My problem is that I don't know how to apply this implementation on an array.
For example if I got the array a:[Int] = [3,1,2].
I can't check if the implementation is working by print(quickSort(a)), because the return type of quickSort is void.
Of course I can't apply quickSort on that array like a.quickSort(a)
I really don't want to change my implementation of the algorithm by much if it isn't the cause of that problem (e.g. just the signature or return type)

Answer 1

Just improve your syntax:
func swap(array: inout [Int], x:Int, y:Int) { //swap elements in array
let temp = array[x]
array[x] = array[y]
array[y] = temp
}
func split(array: inout [Int], u:Int, o:Int , p:Int) -> Int { //where the sorting happens
print(ar, u , o , p)
let pivot:Int = array[p]
swap(array: &array, x: p, y: o)
var pn:Int = u
for j in u..<o {
if(array[j] <= pivot) {
swap(array: &array, x: pn, y: j)
pn = pn+1
}
}
swap(array: &array, x: pn, y: o);
return pn;
}
func quickSortR(array: inout [Int],u:Int ,o:Int) { //recursive call
if(u < o){
let p:Int = o
let pn:Int = split(array: &array, u: u, o: o, p: p)
quickSortR(array: &array,u: u,o: pn-1)
quickSortR(array: &array,u: pn+1,o: o)
}
}
func quickSort(array: inout [Int]) { //initial call
quickSortR(array: &array, u: 0, o: array.count-1)
}
If you use func swap(array: [Int]) array is immutable inside the method. It just copies. Using inout resolves this problem.
To check the code use somethink like this:
var ar = [1]
quickSort(array: &ar)
print(ar)