How to truncate decimals to x places in Swift

Question

I have a really long decimal number (say 17.9384693864596069567) and I want to truncate the decimal to a few decimal places (so I want the output to be 17.9384). I do not want to round the number to 17.9385.

How can I do this?

Answer 1

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))

Answer 2

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.