How to create a pointer in Swift? - swift

Question

I'm working with Swift 3.
I would like to have this C syntax :
int myVar;
int *pointer = &myVar;
So modifying pointer or myVar does the same exact same thing.
Also I don't know if it makes any difference, but in my case myVar is an array containing elements of a class and pointer is a pointer to one element of this array.

Answer 1

The & also exists in Swift but can only be used as part of a parameter list (e.g. init, func, closure).
var i = 5
let ptr = UnsafeMutablePointer(&i)
print(ptr.pointee) // 5
// or
let ptr = UnsafeMutablePointer<Int>.allocate(capacity: 1)
ptr.initialize(to: 5)
// or with a closure
let ptr: UnsafePointer = { $0 }(&i)

Answer 2

(Assuming I understand what you're asking for....)
Try the following code in a playground. It should print "99" three times.
class Row {
var rowNumber = 0
}
var rows = [Row]()
let testRow = Row()
testRow.rowNumber = 1
rows.append(testRow)
let selectedRow = rows[0]
selectedRow.rowNumber = 99
print(testRow.rowNumber)
print(selectedRow.rowNumber)
print(rows[0].rowNumber)
By default, there's no copying of objects as part of an assignment statement. If it were a struct, that would be different.
Adding a bit for completeness:
If you want a similar effect with scalar values instead of objects, Swift supplies various types of wrappers.
let intPointer = UnsafeMutablePointer<Int>.allocate(capacity: 8) // Should be 1, not 8 according to comment re: docs
let other = intPointer
other.pointee = 34
print(intPointer.pointee)
(Warning: I haven't used these wrappers for anything except experimenting in a playground. Don't trust it without doing some research.)

Answer 3

Same example as #Phillip. But I used struct. In this example rows[0] won't change:
struct Row {
var rowNumber = 0
}
var rows = [Row]()
var testRow = Row()
testRow.rowNumber = 1
rows.append(testRow)
var selectedRow = rows[0]
selectedRow.rowNumber = 99
print(testRow.rowNumber) // prints 1
print(selectedRow.rowNumber) // prints 99
print(rows[0].rowNumber) // prints 1

Answer 4

There are no C style pointers (Unsafe Pointer) as the question asks however objects are shared by reference and structures are by value:
Swift assign, pass and return a value by reference for reference type and by copy for Value Type
structures are always copied when they are passed around in your code, but classes are passed by reference.
For example
How to have pointers/ references to objects
class Song {
init(title: String, image: String, file: String, volume: Float, queuePlayer: AVQueuePlayer, playerLooper: AVPlayerLooper?) {
self.title = title
self.image = image
...
}
var title: String
var image: String
...
}
var aSong = Song(title: "", image: "", ...)
var arrOfSongReferences: [Song] = [Song]()
arrOfSongReferences.append(aSong)
var ptrToASong: Song = aSong
aSong = nil
// Due to Swift garbage collection ARC (Automatic Reference Counting), we still have references to the original aSong object so it won't be deleted
If data is struct you cannot do this
struct Song {
var title: String
var image: String
...
}
var aSong: Song = Song(title: "", image: "", ...)
var copyOfASong: Song = aSong
Method
You can also pass by reference into a function
// this would be inside a class, perhaps Player. It doesn't have to be a static btw
static func playSound(_ sound: inout Song, volume: Float = 0.0) {
if (sound.playerLooper == nil) {
...
}
}
// usage
Player.playSound(sound: &aSong)