How to convert Swift optional NSNumber to optional Int? (any improvements on my code?)

Question

What would be the shortest/cleanest way to convert an Optional Number to an Optional Int in Swift?

Is there a better way than this? (see below)

let orderNumberInt : Int?
if event.orderNum != nil {
    orderNumberInt = Int(event.orderNum!)
} else {
    orderNumberInt = nil
}

Bonds

Answer 1

I think most easiest way is

var orderNumberInt = orderNum?.intValue

Also, you can do it like this

var orderNum:NSNumber? = NSNumber(int: 12)
var orderNumberInt:Int? = (orderNum != nil) ? Int(orderNum!) : nil
print(orderNumberInt)

Putrescent

int is the initialiser for Int32 – Puente

thanks guys - you made it hard to choose which was correct first :) But what I'm after specifically here would be "var orderNumberInt = orderNum?.integerValue" so I choose this one, hope thats ok – Bonds

FYI, this method has been renamed from integerValue to intValue in Swift 4. – Twinkling

Answer 2

let number:NSNumber? = NSNumber(integer: 125)


if let integerValue = number?.integerValue {
    print(integerValue)
}

let integerValue = number?.integerValue ?? 0

Puente

Answer 3

orderNumberInt = orderNum?.intValue

This is the best way to do it in Swift.

Reference: https://developer.apple.com/documentation/foundation/nsnumber/1412554-intvalue

Chui

Answer 4

You can use if let syntax

    var number: NSNumber?
    if let value = number as? Int {
        // get int value
    } else {

    }

Bipartite

Answer 5

I am using this simple conversion:

// lets consider following vars
var nsNum: NSNumber?
var oiNum: Int? // oi - optional Int

NSNumber to optional Int?

nsNum = NSNumber.init(value: 18)
oiNum = nsNum as? Int

optional Int? to NSNumber

oiNum = 19
nsNum = oiNum as NSNumber?

p.s. I am new to swift, so if something wrong, feel free to correct :-)

Selves