如何舍入一个双到最近的Int在迅速?

我试图使一个计算器的增长率(双)，将结果四舍五入到最近的整数，并从那里重新计算，如:

let firstUsers = 10.0
let growth = 0.1
var users = firstUsers
var week = 0


while users < 14 {
    println("week \(week) has \(users) users")
    users += users * growth
    week += 1
}

但到目前为止我还不能。

编辑我是这样做的:

var firstUsers = 10.0
let growth = 0.1
var users:Int = Int(firstUsers)
var week = 0


while users <= 14 {
    println("week \(week) has \(users) users")
    firstUsers += firstUsers * growth
    users = Int(firstUsers)
    week += 1
}

虽然我不介意它总是四舍五入，但我不喜欢它，因为firstUsers必须成为一个变量，并在整个程序中改变(为了进行下一次计算)，这是我不希望发生的。

当前回答

一个非常简单的解决方案对我很有效:

  if (62 % 50 != 0) {
      var number = 62 / 50 + 1 // adding 1 is doing the actual "round up"
  }

数字包含值2

2019-11-27 19:36:28

其他回答

这是我想到的最简单的方法

let weightValue = 143.69
print("Rounded off weight value = \(Int(weightValue))")

替代方法:

print("Rounded off weight value = \((String(format: "%.0f", sender.value)))")

2022-02-14 06:17:21

一个非常简单的解决方案对我很有效:

  if (62 % 50 != 0) {
      var number = 62 / 50 + 1 // adding 1 is doing the actual "round up"
  }

数字包含值2

2019-11-27 19:36:28

**In Swift**

var a = 14.123456789
var b = 14.123456789
var c = 14.123456789
var d = 14.123456789
var e = 14.123456789
var f = 14.123456789

a.rounded(.up)                      //15
b.rounded(.down)                    //14
c.rounded(.awayFromZero)            //15
d.rounded(.towardZero)              //14
e.rounded(.toNearestOrAwayFromZero) //14
f.rounded(.toNearestOrEven)         //14

2018-09-26 05:15:20

斯威夫特3: 如果你想四舍五入到某个数字，例如5.678434 -> 5.68，你可以将round()或roundf()函数与乘法组合:

let value:Float = 5.678434
let roundedValue = roundf(value * 100) / 100
print(roundedValue) //5.68

2017-09-20 13:40:21

你也可以扩展FloatingPoint在Swift 3如下:

extension FloatingPoint {
    func rounded(to n: Int) -> Self {
        let n = Self(n)
        return (self / n).rounded() * n

    }
}

324.0.rounded(to: 5)   // 325

2017-01-14 02:34:31

如何舍入一个双到最近的Int在迅速?

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