Swift 3 : Decimal to Int - ios

Question

I tried to convert Decimal to Int with the follow code:
Int(pow(Decimal(size), 2) - 1)
But I get:
.swift:254:43: Cannot invoke initializer for type 'Int' with an argument list of type '(Decimal)'
Here I know pow is returning a Decimal but it seems that Int has no constructors and member functions to convert Decimal to Int.
How can I convert Decimal to Int in Swift 3?

Answer 1

This is my updated answer (thanks to Martin R and the OP for the remarks). The OP's problem was just casting the pow(x: Decimal,y: Int) -> Decimal function to an Int after subtracting 1 from the result. I have answered the question with the help of this SO post for NSDecimal and Apple's documentation on Decimal. You have to convert your result to an NSDecimalNumber, which can in turn be casted into an Int:
let size = Decimal(2)
let test = pow(size, 2) - 1
let result = NSDecimalNumber(decimal: test)
print(Int(result)) // testing the cast to Int

Answer 2

let decimalToInt = (yourDecimal as NSDecimalNumber).intValue
or as #MartinR suggested:
let decimalToInt = NSDecimalNumber(decimal: yourDecimal).intValue

Answer 3

If you have a very long decimal, then beware of rounding errors
let decimal = Decimal(floatLiteral: 100.123456)
let intValue = (decimal as NSDecimalNumber).intValue // This is 100
However
let veryLargeDecimal = Decimal(floatLiteral: 100.123456789123)
let intValue = (veryLargeDecimal as NSDecimalNumber).intValue // This is -84 !
I ensured I rounded my Decimal before I converted it to an Int, using NSDecimalRound (which you can put in an extension of Decimal).
var veryLargeDecimal = Decimal(floatLiteral: 100.123456789123)
var rounded = Decimal()
NSDecimalRound(&rounded, &veryLargeDecimal, 0, .down)
let intValue = (rounded as NSDecimalNumber).intValue // This is now 100

Answer 4

There is nothing wrong with either of the posted answers, but I would like to offer up an extension that reduces the verbosity for scenarios where you need to use this frequently.
extension Decimal {
var int: Int {
return NSDecimalNumber(decimal: self).intValue
}
}
To call it:
let powerDecimal = pow(2, 2) // Output is Decimal
let powerInt = powerDecimal.int // Output is now an Int

Answer 5

Unfortunately there is an intermittent failure using some of the methods provided.
NSDecimalNumber(decimal: <num>).intValue can produce unexpected results...
(lldb) po NSDecimalNumber(decimal: self)
10.6666666666666666666666666666666666666
(lldb) po NSDecimalNumber(decimal: self).intValue
0
I think there is more of a discussion on it here, and #Martin was pointing it out here
Instead of using the decimal value directly, I made a work around that converts the decimal to a whole number before converting the Decimal to an Int.
extension Decimal {
func rounded(_ roundingMode: NSDecimalNumber.RoundingMode = .down, scale: Int = 0) -> Self {
var result = Self()
var number = self
NSDecimalRound(&result, &number, scale, roundingMode)
return result
}
var whole: Self { rounded( self < 0 ? .up : .down) }
var fraction: Self { self - whole }
var int: Int {
NSDecimalNumber(decimal: whole).intValue
}
}

Answer 6

Just use the description of Decimal, String replacement the NSDecimalNumber to bridge it.
extension Decimal {
var intVal: Int? {
return Int(self.description)
}
}