Rounding in Swift with round()

Question

While playing around, I found the round() function in swift. It can be used as below:

round(0.8)

Which will return 1, as expected. Here\'s

Answer 1

This will round to any value not limited by powers of 10.

extension Double {
    func roundToNearestValue(value: Double) -> Double {
        let remainder = self % value
        let shouldRoundUp = remainder >= value/2 ? true : false
        let multiple = floor(self / value)
        let returnValue = !shouldRoundUp ? value * multiple : value * multiple + value
        return returnValue
    }
}

Answer 2

func round(value: Float, decimalPlaces: UInt) {
  decimalValue = pow(10, decimalPlaces)
  round(value * decimalValue) / decimalValue
}
…
func round(value: CGFloat, decimalPlaces: UInt)
func round(value: Double, decimalPlaces: UInt)
func roundf(value: Float, decimalPlaces: UInt)

Answer 3

You can do:

round(1000 * x) / 1000

Answer 4

Updated answer

The round(someDecimal) is the old C style. As of Swift 3, doubles and floats have a built in Swift function.

var x = 0.8
x.round() // x is 1.0 (rounds x in place)

or

var x = 0.8
var y = x.rounded() // y is 1.0, x is 0.8

See my answer fuller answer here (or here) for more details about how different rounding rules can be used.

As other answers have noted, if you want to round to the thousandth, then multiply temporarily by 1000 before you round.

Answer 5

Take a look at Apple's documentation for round() and rounded(_:).

Answer 6

Swift 4:

(x/1000).rounded()*1000