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.

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.

我会用

print(String(format: "%.3f", totalWorkTimeInHours))

然后把。3f改成任何你需要的十进制数

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))
    }

}

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

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