How to truncate decimals to x places in Swift

I figured this one out.

Just floor (round down) the number, with some fancy tricks.

let x = 1.23556789
let y = Double(floor(10000*x)/10000) // leaves on first four decimal places
let z = Double(floor(1000*x)/1000) // leaves on first three decimal places
print(y) // 1.2355
print(z) // 1.235

So, multiply by 1 and the number of 0s being the decimal places you want, floor that, and divide it by what you multiplied it by. And voila.

You can tidy this up even more by making it an extension of Double:

extension Double {
    func truncate(places : Int)-> Double {
        return Double(floor(pow(10.0, Double(places)) * self)/pow(10.0, Double(places)))
    }
}

You use it like this:

var num = 1.23456789
// return the number truncated to 2 places
print(num.truncate(places: 2))

// return the number truncated to 6 places
print(num.truncate(places: 6))