Swift 2扩展

一个更通用的解决方案是以下扩展，适用于Swift 2和iOS 9:

extension Double {
    /// Rounds the double to decimal places value
    func roundToPlaces(places:Int) -> Double {
        let divisor = pow(10.0, Double(places))
        return round(self * divisor) / divisor
    }
}

Swift 3扩展

在Swift 3中，round被圆润取代:

extension Double {
    /// Rounds the double to decimal places value
    func rounded(toPlaces places:Int) -> Double {
        let divisor = pow(10.0, Double(places))
        return (self * divisor).rounded() / divisor
    }
}

返回Double四舍五入到小数点后4位的示例:

let x = Double(0.123456789).roundToPlaces(4)  // x becomes 0.1235 under Swift 2
let x = Double(0.123456789).rounded(toPlaces: 4)  // Swift 3 version

这是一个更灵活的算法舍入到N位有效数字

Swift 3解决方案

extension Double {
// Rounds the double to 'places' significant digits
  func roundTo(places:Int) -> Double {
    guard self != 0.0 else {
        return 0
    }
    let divisor = pow(10.0, Double(places) - ceil(log10(fabs(self))))
    return (self * divisor).rounded() / divisor
  }
}


// Double(0.123456789).roundTo(places: 2) = 0.12
// Double(1.23456789).roundTo(places: 2) = 1.2
// Double(1234.56789).roundTo(places: 2) = 1200

基于Yogi的回答，这里有一个Swift函数来完成这项工作:

func roundToPlaces(value:Double, places:Int) -> Double {
    let divisor = pow(10.0, Double(places))
    return round(value * divisor) / divisor
}

斯威夫特5

使用字符串方法

var yourDouble = 3.12345
//to round this to 2 decimal spaces i could turn it into string
let roundingString = String(format: "%.2f", myDouble)
let roundedDouble = Double(roundingString) //and than back to double
// result is 3.12 

但是使用扩展更容易被接受

extension Double {
    func round(to decimalPlaces: Int) -> Double {
        let precisionNumber = pow(10,Double(decimalPlaces))
        var n = self // self is a current value of the Double that you will round
        n = n * precisionNumber
        n.round()
        n = n / precisionNumber
        return n
    }
}

然后你可以使用:

yourDouble.round(to:2)

Swift 2扩展

一个更通用的解决方案是以下扩展，适用于Swift 2和iOS 9:

extension Double {
    /// Rounds the double to decimal places value
    func roundToPlaces(places:Int) -> Double {
        let divisor = pow(10.0, Double(places))
        return round(self * divisor) / divisor
    }
}

Swift 3扩展

在Swift 3中，round被圆润取代:

extension Double {
    /// Rounds the double to decimal places value
    func rounded(toPlaces places:Int) -> Double {
        let divisor = pow(10.0, Double(places))
        return (self * divisor).rounded() / divisor
    }
}

返回Double四舍五入到小数点后4位的示例:

let x = Double(0.123456789).roundToPlaces(4)  // x becomes 0.1235 under Swift 2
let x = Double(0.123456789).rounded(toPlaces: 4)  // Swift 3 version

var n = 123.111222333
n = Double(Int(n * 10.0)) / 10.0

结果:n = 123.1

将10.0(小数点后1位)更改为100.0(小数点后2位)、1000.0(小数点后3位)中的任意一个，以此作为您想要的小数后的位数。