如何在swift中将double转换为int

Question

所以我想弄清楚如何在一个整数之后让我的程序失去.0时我不需要任何小数位.

@IBOutlet weak var numberOfPeopleTextField: UITextField!
@IBOutlet weak var amountOfTurkeyLabel: UILabel!
@IBOutlet weak var cookTimeLabel: UILabel!
@IBOutlet weak var thawTimeLabel: UILabel!


var turkeyPerPerson = 1.5
var hours: Int = 0
var minutes = 0.0

func multiply (#a: Double, b: Double) -> Double {
    return a * b
}

func divide (a: Double , b: Double) -> Double {
    return a / b
}

@IBAction func calculateButton(sender: AnyObject) {
    var numberOfPeople = numberOfPeopleTextField.text.toInt()
    var amountOfTurkey = multiply(a: 1.5, b: Double(numberOfPeople!))
    var x: Double = amountOfTurkey
    var b: String = String(format:"%.1f", x)
    amountOfTurkeyLabel.text = "Amount of Turkey: " + b + "lbs"

    var time = multiply(a: 15, b:  amountOfTurkey)
    var x2: Double = time
    var b2: String = String(format:"%.1f", x2)
    if (time >= 60) {
        time = time - 60
        hours = hours + 1
        minutes = time
        var hours2: String = String(hours)
        var minutes2: String = String(format: "%.1f", minutes)

        cookTimeLabel.text = "Cook Time: " + hours2 + " hours and " + minutes2 + " minutes"
    }else {
        cookTimeLabel.text = "Cook Time: " + b2 + "minutes"
    }

}

}

我是否需要制作一个if语句以某种方式将Double转换为Int以使其工作？

Answer 1

您可以使用:

Int(yourDoubleValue)

这会将double转换为int.

或者当您使用String格式时使用0而不是1:

String(format: "%.0f", yourDoubleValue)

这将只显示没有小数位的Double值,而不将其转换为int.

Answer 2

最好Double在转换之前验证值的大小,否则可能会崩溃.

extension Double {
    func toInt() -> Int? {
        if self >= Double(Int.min) && self < Double(Int.max) {
            return Int(self)
        } else {
            return nil
        }
    }
}

崩溃很容易证明,只需使用Int(Double.greatestFiniteMagnitude).

Answer 3

那应该有效：

// round your double so that it will be exactly-convertible
if let converted = Int(exactly: double.rounded()) {
  doSomethingWithInteger(converted)
} else {
  // double wasn't convertible for a reason, it probably overflows
  reportAnError("\(double) is not convertible")
}

init(exactly:)与 几乎相同init(:)，唯一的区别是init(exactly:)不会崩溃，但可能会在失败时init(:)调用。fatalError(:)

你可以在这里检查他们的实现

• init(:)
• init(exactly:)

Answer 4

其他方式：

extension Double {
    
    func toInt() -> Int? {
        let roundedValue = rounded(.toNearestOrEven)
        return Int(exactly: roundedValue)
    }
    
}

Answer 5

抑制（仅）Double值的字符串表示中的 .0 小数位的更通用方法是使用NSNumberFormatter. 它还考虑当前语言环境的数字格式。

let x : Double = 2.0
let doubleAsString = NumberFormatter.localizedString(from: (NSNumber(value: x), numberStyle: .decimal) 
// --> "2"