Display Double as Int if it is a whole number in Swift

Question

I am trying to write a method that will check to see if the Double ends in .0 vs anything else. So for example checking if it is 2.0 vs 2.12345.

If it is 2.0 then I want it to display just 2, but if it is 2.12345 then show 2.12345.

I had the idea of writing a func that took the Double as a parameter and returned an Int, but if the input turns out to be .12345 it cannot return it.

So I was wondering if the way I was thinking would work with tweaking or if there is a much similar means.

I was thinking maybe the return type could somehow be both Int & Double, but I do not know how that would work.

This is what I have so far

  private func checkIfWholeNumber(displayedDouble: Double) -> Int {
    if displayedDouble.truncatingRemainder(dividingBy: 1) == 0 {
      return Int(displayedDouble)
    } else {
      return //
    }
  }

Answer 1

Convert Double into Number and then to String if u want to show it on label

let i = 1.0
let j = 1.2345

// Value in String

let a = NSNumber(value: j).stringValue  // "1.2345"
let b = NSNumber(value: i).stringValue  // "1" 

// Value in Double

let c = NSNumber(value: i).doubleValue  // 1.0
let d = NSNumber(value: j).doubleValue  // 1.2345

Answer 2

simply use rounded function.

func check(val:Double)->Int?{
    let rounded = val.rounded()
    return rounded == val ? Int(rounded) : nil
}