Rounding in Swift with round()

问题:

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 my question:

how do I round by thousandths in swift?

I want to be able to plug in a number, say 0.6849, and get 0.685 back. How does round() do this? Or, does it not, in which case, what function does?

回答1:

You can do:

round(1000 * x) / 1000

回答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)

回答3:

Swift 3

The round(someDecimal) is the old C style. Doubles and floats have a built in Swift function now.

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.

回答4:

Here's one way to do it. You could easily do this for Float, or probably make it generic so it's for any of those.

public extension CGFloat {
    func roundToDecimals(decimals: Int = 2) -> CGFloat {
        let multiplier = CGFloat(10^decimals)
        return round(multiplier * self) / multiplier
    }
}

回答5:

Swift 4:

(x/1000).rounded()*1000

回答6:

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
    }
}