How to create fixed-size 2d array in swift

Question

Recently I am doing some code tests and found that fixed-size array creation is not as simple in Swift compare to other languages, say c++.

I have seen the solution for 1d array here : link:

One Dimension array, compare c++ with Swift:

// In c++

int array1[64];    // 1-dimension array size 64

// In Swift

var array1 = [Int?](repeating: nil, count: 64) // 1 dimension array size 64

For 2-dimension array:

// in c++

int array2[64][64]; // 2-dimension array size 64x64

// in Swift

var array2 : [[Int?]] = ???????

How to initiate a fixed-size 2d array in swift?

Marco Leong · Answer 1 · 2019-04-19T02:46:17.720

2

I found the answer by testing it out in Playground, here's the solution:

In c++

int array1[64];    // 1-dimension array size 64

int array2[64][64]; // 2-dimension array size 64x64

in Swift:

var array1 = [Int?](repeating: nil, count: 64) // 1 dimension array

var array2 = [[Int?]](
 repeating: [Int?](repeating: nil, count: 64)
 count: 64
) // 2-dimension array size 64x64

// Access it like normal

array2[4][2] = 42
print(array2[4][2]) // output: 42

A bonus, 3-dimension array in swift !!!

var array3 = 
[[[Int?]]](
  repeating: [[Int?]](
    repeating: [Int?](
      repeating: nil,
      count: 3),
  count: 3),
count: 3)

edited Apr 19 '19 at 02:46

answered Apr 19 '19 at 02:37

Marco Leong

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4
11

In what sense is this "fixed size"? And what's `var` in the 5th line? – matt Apr 19 '19 at 02:44
declaring the size of the array without any value in it – Marco Leong Apr 19 '19 at 02:45
just noticed the error of the `var` and corrected it – Marco Leong Apr 19 '19 at 02:46
1

Okay but just so you know it is not “fixed”. The size can change. – matt Apr 19 '19 at 02:50
you are right, i think it is more like a pre-declared and initiated with a given size? what is a better terminology? – Marco Leong Apr 19 '19 at 06:43

How to create fixed-size 2d array in swift

1 Answers1