这在Swift 5中似乎有效。

令人惊讶的是，现在还没有这样的标准函数。

//用四舍五入截断双位数到小数点后n位

extension Double {

    func truncate(to places: Int) -> Double {
    return Double(Int((pow(10, Double(places)) * self).rounded())) / pow(10, Double(places))
    }

}

你可以添加这个扩展:

extension Double {
    var clean: String {
        return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(format: "%.2f", self)
    }
}

像这样叫它:

let ex: Double = 10.123546789
print(ex.clean) // 10.12

如果你想在逗号后面只有0表示圆，试试这个:

extension Double {
    func isInteger() -> Any {
        let check = floor(self) == self
        if check {
            return Int(self)
        } else {
            return self
        }
    }
}

let toInt: Double = 10.0
let stillDouble: Double = 9.12

print(toInt.isInteger) // 10
print(stillDouble.isInteger) // 9.12

这个解决方案对我很有效。XCode 13.3.1 & Swift 5

extension Double {
    
    func rounded(decimalPoint: Int) -> Double {
        let power = pow(10, Double(decimalPoint))
       return (self * power).rounded() / power
    }
}

测试:

print(-87.7183123123.rounded(decimalPoint: 3))
print(-87.7188123123.rounded(decimalPoint: 3))
print(-87.7128123123.rounded(decimalPoint: 3))

结果:

-87.718
-87.719
-87.713

一个方便的方法是使用Double类型的扩展

extension Double {
    var roundTo2f: Double {return Double(round(100 *self)/100)  }
    var roundTo3f: Double {return Double(round(1000*self)/1000) }
}

用法:

let regularPie:  Double = 3.14159
var smallerPie:  Double = regularPie.roundTo3f  // results 3.142
var smallestPie: Double = regularPie.roundTo2f  // results 3.14

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
}