Two-dimensional array in Swift

Question

I get so confused about 2D arrays in Swift. Let me describe step by step. And would you please correct me if I am wrong.

First of all; declaration of an empty array:

class test{
    var my2Darr = Int[][]()
}

Secondly fill the array. (such as my2Darr[i][j] = 0 where i, j are for-loop variables)

class test {
    var my2Darr = Int[][]()
    init() {
        for(var i:Int=0;i<10;i++) {
            for(var j:Int=0;j<10;j++) {
                my2Darr[i][j]=18   /*  Is this correct?  */
            }
        }
    }
}

And Lastly, Editing element of in array

class test {
    var my2Darr = Int[][]()
    init() {
        ....  //same as up code
    }
    func edit(number:Int,index:Int){
        my2Darr[index][index] = number
        // Is this correct? and What if index is bigger
        // than i or j... Can we control that like 
        if (my2Darr[i][j] == nil) { ...  }   */
    }
}

Answer 1

Define mutable array

// 2 dimensional array of arrays of Ints 
var arr = [[Int]]() 

OR:

// 2 dimensional array of arrays of Ints 
var arr: [[Int]] = [] 

OR if you need an array of predefined size (as mentioned by @0x7fffffff in comments):

// 2 dimensional array of arrays of Ints set to 0. Arrays size is 10x5
var arr = Array(count: 3, repeatedValue: Array(count: 2, repeatedValue: 0))

// ...and for Swift 3+:
var arr = Array(repeating: Array(repeating: 0, count: 2), count: 3)

Change element at position

arr[0][1] = 18

OR

let myVar = 18
arr[0][1] = myVar

Change sub array

arr[1] = [123, 456, 789] 

OR

arr[0] += 234

OR

arr[0] += [345, 678]

If you had 3x2 array of 0(zeros) before these changes, now you have:

[
  [0, 0, 234, 345, 678], // 5 elements!
  [123, 456, 789],
  [0, 0]
]

So be aware that sub arrays are mutable and you can redefine initial array that represented matrix.

Examine size/bounds before access

let a = 0
let b = 1

if arr.count > a && arr[a].count > b {
    println(arr[a][b])
}

Remarks: Same markup rules for 3 and N dimensional arrays.