不是斯威夫特，但我相信你明白我的意思。

pow10np = pow(10,num_places);
val = round(val*pow10np) / pow10np;

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))
    }

}

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.

你可以使用Swift的round函数来实现这一点。

要对Double进行3位精度的四舍五入，首先将其乘以1000，四舍五入，然后将四舍五入结果除以1000:

let x = 1.23556789
let y = Double(round(1000 * x) / 1000)
print(y) /// 1.236

与任何类型的printf(…)或String(format:…)解决方案不同，此操作的结果仍然是Double类型。

编辑: 关于它有时不能工作的评论，请阅读这篇:每个程序员都应该知道的关于浮点算术的事情

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

我会用

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

然后把。3f改成任何你需要的十进制数