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

我想知道是否有可能纠正用户的输入。也就是说，如果他们输入三个小数而不是两个小数来表示一美元的金额。比如说1.111而不是1.11，你能通过四舍五入来修复它吗?出于很多原因，答案是否定的!对于金钱，任何超过0.001的东西最终都会在真正的支票簿上产生问题。

下面是一个函数，用于检查用户输入的句点之后是否有太多值。但是它允许1。、1.1和1.11。

假设已经检查了该值，以成功地从String转换为Double。

//func need to be where transactionAmount.text is in scope

func checkDoublesForOnlyTwoDecimalsOrLess()->Bool{


    var theTransactionCharacterMinusThree: Character = "A"
    var theTransactionCharacterMinusTwo: Character = "A"
    var theTransactionCharacterMinusOne: Character = "A"

    var result = false

    var periodCharacter:Character = "."


    var myCopyString = transactionAmount.text!

    if myCopyString.containsString(".") {

         if( myCopyString.characters.count >= 3){
                        theTransactionCharacterMinusThree = myCopyString[myCopyString.endIndex.advancedBy(-3)]
         }

        if( myCopyString.characters.count >= 2){
            theTransactionCharacterMinusTwo = myCopyString[myCopyString.endIndex.advancedBy(-2)]
        }

        if( myCopyString.characters.count > 1){
            theTransactionCharacterMinusOne = myCopyString[myCopyString.endIndex.advancedBy(-1)]
        }


          if  theTransactionCharacterMinusThree  == periodCharacter {

                            result = true
          }


        if theTransactionCharacterMinusTwo == periodCharacter {

            result = true
        }



        if theTransactionCharacterMinusOne == periodCharacter {

            result = true
        }

    }else {

        //if there is no period and it is a valid double it is good          
        result = true

    }

    return result


}

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.

格式化double属性的最好方法是使用Apple预定义的方法。

mutating func round(_ rule: FloatingPointRoundingRule)

FloatingPointRoundingRule是一个枚举，有以下几种可能

枚举的案例:

案例awayFromZero 四舍五入到最接近的允许值，其大小大于或等于源的大小。

情况下 四舍五入到小于或等于源的最接近的允许值。

案例toNearestOrAwayFromZero 四舍五入到最接近的允许值;如果两个值相等接近，则选择大小较大的值。

案例toNearestOrEven 四舍五入到最接近的允许值;如果两个值相等接近，则选择偶数。

案例towardZero 四舍五入到最接近的允许值，其大小小于或等于源的大小。

情况下了 四舍五入到最接近的允许值，该值大于或等于源。

var aNumber : Double = 5.2
aNumber.rounded(.up) // 6.0

为了避免Float不完美，请使用Decimal

extension Float {
    func rounded(rule: NSDecimalNumber.RoundingMode, scale: Int) -> Float {
        var result: Decimal = 0
        var decimalSelf = NSNumber(value: self).decimalValue
        NSDecimalRound(&result, &decimalSelf, scale, rule)
        return (result as NSNumber).floatValue
    }
}

前女友。 1075.58在使用Float时四舍五入为1075.57 1075.58在使用十进制时四舍五入为1075.58，比例为2和。down

当我打印totalWorkTimeInHours时，我如何将其四舍五入到1.543 ?

要将totalWorkTimeInHours四舍五入为3位数字以便打印，使用String构造函数，它接受一个格式字符串:

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