使用内置的达尔文基金会图书馆

斯威夫特3

extension Double {
    func round(to places: Int) -> Double {
        let divisor = pow(10.0, Double(places))
        return Darwin.round(self * divisor) / divisor
    }
}

用法:

let number:Double = 12.987654321
print(number.round(to: 3)) 

输出:12.988

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。

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
}

Lots of example are using maths, the problem is floats are approximations of real number, there is no way to express 0.1 (1/10) exactly as a float just as there is no exact way to express ⅓ exactly using decimal points, so you need to ask your self exactly what your are trying to achieve, if you just want to display them leave them as they are in code, trying to round them is going to justify give you less accurate result as you are throwing away precious, round ⅓ in decimal notation to 1 decimal place is not going to give you a number closer to ⅓, us NumberFormate to round it, if you have something like a viewModel class it can be used to return a string representation to your models numbers. NumberFormaters give you lots of control on how numbers are formatted and the number of decimal places you want.

Swift最佳方式

这就是我把四舍五入到小数点后两位的方法

let val = round(100 * scale) / 100

为了方便使用，我创建了一个扩展:

extension Double {
    var threeDigits: Double {
        return (self * 1000).rounded(.toNearestOrEven) / 1000
    }
    
    var twoDigits: Double {
        return (self * 100).rounded(.toNearestOrEven) / 100
    }
    
    var oneDigit: Double {
        return (self * 10).rounded(.toNearestOrEven) / 10
    }
}

var myDouble = 0.12345
print(myDouble.threeDigits)
print(myDouble.twoDigits)
print(myDouble.oneDigit)

打印结果如下:

0.123
0.12
0.1

感谢其他答案的启发!