How to create 3D NSMutableArray - ios

Question

How can I create a 3D Mutable Array of Int?
I want to be able to insert an element by saying arr[4][3][5] = 4, and not have to determine the size of the array beforehand.

Answer 1

Discussion (see below for final code)
Here is a way to do it: First we extend _ArrayType with a subscript that automatically extends the array when accessing an index out of bounds. We will the array with a given fallback value:
extension _ArrayType {
subscript (extending pos: Int, fallback fallback: Generator.Element) -> Generator.Element {
mutating get {
while count <= pos { append(fallback) }
return self[pos]
}
mutating set {
while count <= pos { append(fallback) }
self[pos] = newValue
}
}
}
This is useable like this:
var a : [Int] = []
a[extending: 3, fallback: 0] = 10
print(a) // [0, 0, 0, 10]
Now we can extend the 3-dimensional arrays to have a convenient subscript that auto-extends like this:
extension _ArrayType where
Generator.Element : _ArrayType,
Generator.Element.Generator.Element : _ArrayType {
typealias T = Generator.Element.Generator.Element.Generator.Element
subscript (x: Int, y: Int, z: Int, `default`: T) -> T {
mutating get {
return self[extending: x, fallback: []][extending: y, fallback: []][extending: z, fallback: `default`]
}
mutating set {
self[extending: x, fallback: []][extending: y, fallback: []][extending: z, fallback: `default`] = newValue
}
}
}
Usage:
var array : [[[Int?]]] = []
array[1, 1, 0, nil] = 10
array[2, 0, 1, nil] = 4
print(array) // [[], [[], [Optional(10)]], [[nil, Optional(4)]]]
Now if you want it even more convenient by telling beforehand what the default value should be for a certain type you can create a HasDefault protocol:
protocol HasDefault {
static var `default` : Self { get }
}
And extend your types with it:
extension Int : HasDefault {
static var `default` = 0
}
extension Bool : HasDefault {
static var `default` = false
}
Also make a subscript that doesn't take the last argument:
extension _ArrayType where
Generator.Element : _ArrayType,
Generator.Element.Generator.Element : _ArrayType,
Generator.Element.Generator.Element.Generator.Element : HasDefault {
typealias E = Generator.Element.Generator.Element.Generator.Element
subscript (x: Int, y: Int, z: Int) -> E {
mutating get {
return self[x, y, z, E.`default`]
}
mutating set {
self[x, y, z, E.`default`] = newValue
}
}
}
Now this is useable the way you want it:
var array : [[[Int]]] = []
array[0, 1, 0] = 3
array[1, 2, 3] = 10
print(array[2, 0, 1]) // 0
print(array) // [[[], [3]], [[], [], [0, 0, 0, 10]], [[0, 0]]]
Unfortunately it's not (yet) possible to extend Optional with HasDefault (which would be nil), but you can make a custom subscript for NilLiteralConvertibles which Optional conforms to:
extension _ArrayType where
Generator.Element : _ArrayType,
Generator.Element.Generator.Element : _ArrayType,
Generator.Element.Generator.Element.Generator.Element : NilLiteralConvertible {
subscript (x: Int, y: Int, z: Int) -> Generator.Element.Generator.Element.Generator.Element {
mutating get {
return self[x, y, z, nil]
}
mutating set {
self[x, y, z, nil] = newValue
}
}
}
Use like this:
var array : [[[Int?]]] = []
array[0, 1, 0] = 3
array[1, 2, 3] = 10
print(array[2, 0, 1]) // nil
print(array) // [[[], [3]], [[], [], [nil, nil, nil, 10]], [[nil, nil]]]
I heavily recommend this approach, since the whole purpose of Optionals is to represent no value.
TL;DR / The final code
extension _ArrayType {
subscript (extending pos: Int, fallback fallback: Generator.Element) -> Generator.Element {
mutating get {
while count <= pos { append(fallback) }
return self[pos]
}
mutating set {
while count <= pos { append(fallback) }
self[pos] = newValue
}
}
}
extension _ArrayType where
Generator.Element : _ArrayType,
Generator.Element.Generator.Element : _ArrayType,
Generator.Element.Generator.Element.Generator.Element : NilLiteralConvertible {
subscript (x: Int, y: Int, z: Int) -> Generator.Element.Generator.Element.Generator.Element {
mutating get {
return self[extending: x, fallback: []][extending: y, fallback: []][extending: z, fallback: nil]
}
mutating set {
self[extending: x, fallback: []][extending: y, fallback: []][extending: z, fallback: nil] = newValue
}
}
}
var array : [[[Int?]]] = []
array[0, 1, 0] = 3
array[1, 2, 3] = 10
print(array[2, 0, 1]) // nil
print(array) // [[[], [3]], [[], [], [nil, nil, nil, 10]], [[nil, nil]]]

Answer 2

var z = [Int](count:10, repeatedValue: 0)
var y = [[Int]](count:10, repeatedValue: z)
var x = [[[Int]]](count:10, repeatedValue: y)
x[A][B][C] = 4
var val = x[2][1][1]
Swift 3 in Xcode 8 beta 2 playground
var z = [Int](repeatElement(0, count: 10))
var y = [[Int]](repeatElement(z, count: 10))
var x = [[[Int]]](repeatElement(y, count: 10))
Possible answer
var z = [0]
var y = [z]
var x = [y]
x[0][0].append(2)
x[0].append([1])
x.append([[3,5],[0]])
print(x)
x[0][0][0] = 2
print(x)
//Output
[[[0, 2], [1]], [[3, 5], [0]]]
[[[2, 2], [1]], [[3, 5], [0]]]