如何将小数截断到 Swift 中的 x 位
How to truncate decimals to x places in Swift
我有一个很长的小数（比如 17.9384693864596069567），我想将小数截断到小数点后几位（所以我希望输出为 17.9384）。我不想将数字四舍五入为17.9385。
我该怎么做？
我想通了。
用一些花哨的技巧将数字向下舍入（向下舍入）。
let x = 1.23556789
let y = Double(floor(10000*x)/10000) // leaves on first four decimal places
let z = Double(floor(1000*x)/1000) // leaves on first three decimal places
print(y) // 1.2355
print(z) // 1.235
所以，乘以 1 和你想要的小数位数 0，将其取底，然后除以你乘以的值。瞧。
您可以通过将其作为 Double 的扩展来进一步整理它：
extension Double {
func truncate(places : Int)-> Double {
return Double(floor(pow(10.0, Double(places)) * self)/pow(10.0, Double(places)))
}
}
你这样使用它：
var num = 1.23456789
// return the number truncated to 2 places
print(num.truncate(places: 2))
// return the number truncated to 6 places
print(num.truncate(places: 6))
小数点后特定数字的代码是：
let number = 17.9384693864596069567;
let merichle = Float(String(format: "%.1f", (number * 10000)).dropLast(2))!/10000
//merichle = 17.9384
最终，您的号码会被截断而没有舍入...
Drag/Drop解，IOS，Swift4
将此代码复制到您的应用程序中...
import Foundation
func truncateDigitsAfterDecimal(number: Double, afterDecimalDigits: Int) -> Double {
if afterDecimalDigits < 1 || afterDecimalDigits > 512 {return 0.0}
return Double(String(format: "%.\(afterDecimalDigits)f", number))!
}
然后你可以这样调用这个函数：
truncateDigitsAfterDecimal(number: 45.123456789, afterDecimalDigits: 3)
将产生以下内容：
45.123
extension Double {
/// Rounds the double to decimal places value
func roundToPlaces(_ places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
func cutOffDecimalsAfter(_ places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self*divisor).rounded(.towardZero) / divisor
}
}
let a:Double = 1.228923598
print(a.roundToPlaces(2)) // 1.23
print(a.cutOffDecimalsAfter(2)) // 1.22
斯威夫特用户界面：
如果您要在视图中截断以格式化输出，而不是计算，SwiftUI 包含一种方便的方法来使用 C 格式说明符作为 Text() 签名的一部分。
import SwiftUI
let myDouble = 17.93846938645960695
Text("\(myDouble, specifier: "%.2f")")
///Device display: 17.94
以上代码会将 Double 的内容直接输出到 View，正确四舍五入到百位，但保留 Double 的值以供进一步计算。
如果作为 SwiftUI 的用户，您不熟悉 C 语言格式说明符，此 link 包含有用的信息：https://en.wikipedia.org/wiki/Printf_format_string
您可以通过以下方式保持简单：
String(format: "%.0f", ratio*100)
其中 0 是您要允许的小数位数。在这种情况下为零。比率是双倍的：0.5556633。
希望能帮助到你。
在 swift5 中也可以通过创建十进制扩展来截断为十进制
extension Decimal {
func truncation(by digit: Int) -> Decimal {
var initialDecimal = self
var roundedDecimal = Decimal()
NSDecimalRound(&roundedDecimal, &initialDecimal, digit, .plain)
return roundedDecimal
}
十进制值的用例
value = Decimal(2.56430).truncation(by:2)
value = 2.560000（截断后）
回答 Swift 5.2
我看了很多答案，截断时总是遇到转换问题。根据我的数学知识，通过截断我知道如果我有 3.1239 并且我想要 3 位小数那么我将有 3.123 没有四舍五入 (!= 3.1234).
也许，由于过程的性质，我在 Doubles 上总是成功，但在 Floats 上总是有问题。
我的方法是创建 BinaryFloatingPoint 的扩展，以便我可以将它重新用于 Float、CGFLoat、Double ...
以下扩展获取 BinaryFloatingPoint 并且可以 return String 或 BinaryFloatingPoint 值与给定的 numberOfDecimals 并且它处理不同类型的情况：
extension Numeric where Self: BinaryFloatingPoint {
/// Retruns the string value of the BinaryFloatingPoint. The initiaiser
var toString: String {
return String(describing: self)
}
/// Returns the number of decimals. It will be always greater than 0
var numberOfDecimals: Int {
return toString.count - String(Int(self)).count - 1
}
/// Returns a Number with a certain number of decimals
/// - Parameters:
/// - Parameter numberOfDecimals: Number of decimals to return
/// - Returns: BinaryFloatingPoint with number of decimals especified
func with(numberOfDecimals: Int) -> Self {
let stringValue = string(numberOfDecimals: numberOfDecimals)
if self is Double {
return Double(stringValue) as! Self
} else {
return Float(stringValue) as! Self
}
}
/// Returns a string representation with a number of decimals
/// - Parameters:
/// - Parameter numberOfDecimals: Number of decimals to return
/// - Returns: String with number of decimals especified
func string(numberOfDecimals: Int) -> String {
let selfString = toString
let selfStringComponents = selfString.components(separatedBy: ".")
let selfStringIntegerPart = selfStringComponents[0]
let selfStringDecimalPart = selfStringComponents[1]
if numberOfDecimals == 0 {
return selfStringIntegerPart
} else {
if selfStringDecimalPart.count == numberOfDecimals {
return [selfStringIntegerPart,
selfStringDecimalPart].joined(separator: ".")
} else {
if selfStringDecimalPart.count > numberOfDecimals {
return [selfStringIntegerPart,
String(selfStringDecimalPart.prefix(numberOfDecimals))].joined(separator: ".")
} else {
let difference = numberOfDecimals - selfStringDecimalPart.count
let addedCharacters = [Character].init(repeating: "0", count: difference)
return [selfStringIntegerPart,
selfStringDecimalPart+addedCharacters].joined(separator: ".")
}
}
}
}
}
它可能看起来很老派，但我的所有测试都通过了：
func test_GivenADecimalNumber_ThenAssertNumberOfDecimalsWanted() {
//No decimals
XCTAssertEqual(Float(3).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Double(3).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 0), 3)
//numberOfDecimals == totalNumberOfDecimals
XCTAssertEqual(Float(3.00).with(numberOfDecimals: 2), 3.00)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 2), 3.09)
XCTAssertEqual(Float(3.01).with(numberOfDecimals: 2), 3.01)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 3), 3.999)
XCTAssertEqual(Float(3.991).with(numberOfDecimals: 3), 3.991)
XCTAssertEqual(Double(3.00).with(numberOfDecimals: 2), 3.00)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 2), 3.09)
XCTAssertEqual(Double(3.01).with(numberOfDecimals: 2), 3.01)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 3), 3.999)
XCTAssertEqual(Double(3.991).with(numberOfDecimals: 3), 3.991)
//numberOfDecimals < totalNumberOfDecimals
XCTAssertEqual(Float(3.00).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Float(3.01).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 2), 3.99)
XCTAssertEqual(Float(3.991).with(numberOfDecimals: 2), 3.99)
XCTAssertEqual(Double(3.00).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Double(3.01).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 2), 3.99)
XCTAssertEqual(Double(3.991).with(numberOfDecimals: 2), 3.99)
//numberOfDecimals > totalNumberOfDecimals
XCTAssertEqual(Float(3.00).with(numberOfDecimals: 3), 3.000)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 3), 3.090)
XCTAssertEqual(Float(3.01).with(numberOfDecimals: 3), 3.010)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 4), 3.9990)
XCTAssertEqual(Float(3.991).with(numberOfDecimals: 4), 3.9910)
XCTAssertEqual(Double(3.00).with(numberOfDecimals: 3), 3.000)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 3), 3.090)
XCTAssertEqual(Double(3.01).with(numberOfDecimals: 3), 3.010)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 4), 3.9990)
XCTAssertEqual(Double(3.991).with(numberOfDecimals: 4), 3.9910)
}
func test_GivenADecimal_ThenAssertStringValueWithDecimalsWanted() {
//No decimals
XCTAssertEqual(Float(3).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Double(3).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 0), "3")
//numberOfDecimals == totalNumberOfDecimals
XCTAssertEqual(Float(3.00).string(numberOfDecimals: 2), "3.00")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 2), "3.09")
XCTAssertEqual(Float(3.01).string(numberOfDecimals: 2), "3.01")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 3), "3.999")
XCTAssertEqual(Float(3.991).string(numberOfDecimals: 3), "3.991")
XCTAssertEqual(Double(3.00).string(numberOfDecimals: 2), "3.00")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 2), "3.09")
XCTAssertEqual(Double(3.01).string(numberOfDecimals: 2), "3.01")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 3), "3.999")
XCTAssertEqual(Double(3.991).string(numberOfDecimals: 3), "3.991")
//numberOfDecimals < totalNumberOfDecimals
XCTAssertEqual(Float(3.00).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Float(3.01).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 2), "3.99")
XCTAssertEqual(Float(3.991).string(numberOfDecimals: 2), "3.99")
XCTAssertEqual(Double(3.00).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Double(3.01).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 2), "3.99")
XCTAssertEqual(Double(3.991).string(numberOfDecimals: 2), "3.99")
//numberOfDecimals > totalNumberOfDecimals
XCTAssertEqual(Float(3.00).string(numberOfDecimals: 3), "3.000")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 3), "3.090")
XCTAssertEqual(Float(3.01).string(numberOfDecimals: 3), "3.010")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 4), "3.9990")
XCTAssertEqual(Float(3.991).string(numberOfDecimals: 4), "3.9910")
XCTAssertEqual(Double(3.00).string(numberOfDecimals: 3), "3.000")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 3), "3.090")
XCTAssertEqual(Double(3.01).string(numberOfDecimals: 3), "3.010")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 4), "3.9990")
XCTAssertEqual(Double(3.991).string(numberOfDecimals: 4), "3.9910")
}
方式一：如果你不想为此创建任何新函数，你可以通过这种方式直接获取舍入值。
var roundedValue = (decimalValue * pow(10.0, Double(numberOfPlaces))).rounded())/pow(10.0, Double(numberOfPlaces)
示例：
var numberOfPlaces = 2
var decimalValue = 13312.2423423523523
print("\(((decimalValue * pow(10.0, Double(numberOfPlaces))).rounded())/pow(10.0, Double(numberOfPlaces)))")
结果：13312.24
方式二：如果只想打印，可以使用：
print(String(format: "%.\(numberOfPlaces)f",decimalValue))
例子
var numberOfPlaces = 4
var decimalValue = 13312.2423423523523
print(String(format: "%.\(numberOfPlaces)f",decimalValue))
我有一个很长的小数（比如 17.9384693864596069567），我想将小数截断到小数点后几位（所以我希望输出为 17.9384）。我不想将数字四舍五入为17.9385。
我该怎么做？
我想通了。
用一些花哨的技巧将数字向下舍入（向下舍入）。
let x = 1.23556789
let y = Double(floor(10000*x)/10000) // leaves on first four decimal places
let z = Double(floor(1000*x)/1000) // leaves on first three decimal places
print(y) // 1.2355
print(z) // 1.235
所以，乘以 1 和你想要的小数位数 0，将其取底，然后除以你乘以的值。瞧。
您可以通过将其作为 Double 的扩展来进一步整理它：
extension Double {
func truncate(places : Int)-> Double {
return Double(floor(pow(10.0, Double(places)) * self)/pow(10.0, Double(places)))
}
}
你这样使用它：
var num = 1.23456789
// return the number truncated to 2 places
print(num.truncate(places: 2))
// return the number truncated to 6 places
print(num.truncate(places: 6))
小数点后特定数字的代码是：
let number = 17.9384693864596069567;
let merichle = Float(String(format: "%.1f", (number * 10000)).dropLast(2))!/10000
//merichle = 17.9384
最终，您的号码会被截断而没有舍入...
Drag/Drop解，IOS，Swift4
将此代码复制到您的应用程序中...
import Foundation
func truncateDigitsAfterDecimal(number: Double, afterDecimalDigits: Int) -> Double {
if afterDecimalDigits < 1 || afterDecimalDigits > 512 {return 0.0}
return Double(String(format: "%.\(afterDecimalDigits)f", number))!
}
然后你可以这样调用这个函数：
truncateDigitsAfterDecimal(number: 45.123456789, afterDecimalDigits: 3)
将产生以下内容：
45.123
extension Double {
/// Rounds the double to decimal places value
func roundToPlaces(_ places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
func cutOffDecimalsAfter(_ places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self*divisor).rounded(.towardZero) / divisor
}
}
let a:Double = 1.228923598
print(a.roundToPlaces(2)) // 1.23
print(a.cutOffDecimalsAfter(2)) // 1.22
斯威夫特用户界面： 如果您要在视图中截断以格式化输出，而不是计算，SwiftUI 包含一种方便的方法来使用 C 格式说明符作为 Text() 签名的一部分。
import SwiftUI
let myDouble = 17.93846938645960695
Text("\(myDouble, specifier: "%.2f")")
///Device display: 17.94
以上代码会将 Double 的内容直接输出到 View，正确四舍五入到百位，但保留 Double 的值以供进一步计算。
如果作为 SwiftUI 的用户，您不熟悉 C 语言格式说明符，此 link 包含有用的信息：https://en.wikipedia.org/wiki/Printf_format_string
您可以通过以下方式保持简单：
String(format: "%.0f", ratio*100)
其中 0 是您要允许的小数位数。在这种情况下为零。比率是双倍的：0.5556633。 希望能帮助到你。
在 swift5 中也可以通过创建十进制扩展来截断为十进制
extension Decimal {
func truncation(by digit: Int) -> Decimal {
var initialDecimal = self
var roundedDecimal = Decimal()
NSDecimalRound(&roundedDecimal, &initialDecimal, digit, .plain)
return roundedDecimal
}
十进制值的用例
value = Decimal(2.56430).truncation(by:2)
value = 2.560000（截断后）
回答 Swift 5.2
我看了很多答案，截断时总是遇到转换问题。根据我的数学知识，通过截断我知道如果我有 3.1239 并且我想要 3 位小数那么我将有 3.123 没有四舍五入 (!= 3.1234).
也许，由于过程的性质，我在 Doubles 上总是成功，但在 Floats 上总是有问题。
我的方法是创建 BinaryFloatingPoint 的扩展，以便我可以将它重新用于 Float、CGFLoat、Double ...
以下扩展获取 BinaryFloatingPoint 并且可以 return String 或 BinaryFloatingPoint 值与给定的 numberOfDecimals 并且它处理不同类型的情况：
extension Numeric where Self: BinaryFloatingPoint {
/// Retruns the string value of the BinaryFloatingPoint. The initiaiser
var toString: String {
return String(describing: self)
}
/// Returns the number of decimals. It will be always greater than 0
var numberOfDecimals: Int {
return toString.count - String(Int(self)).count - 1
}
/// Returns a Number with a certain number of decimals
/// - Parameters:
/// - Parameter numberOfDecimals: Number of decimals to return
/// - Returns: BinaryFloatingPoint with number of decimals especified
func with(numberOfDecimals: Int) -> Self {
let stringValue = string(numberOfDecimals: numberOfDecimals)
if self is Double {
return Double(stringValue) as! Self
} else {
return Float(stringValue) as! Self
}
}
/// Returns a string representation with a number of decimals
/// - Parameters:
/// - Parameter numberOfDecimals: Number of decimals to return
/// - Returns: String with number of decimals especified
func string(numberOfDecimals: Int) -> String {
let selfString = toString
let selfStringComponents = selfString.components(separatedBy: ".")
let selfStringIntegerPart = selfStringComponents[0]
let selfStringDecimalPart = selfStringComponents[1]
if numberOfDecimals == 0 {
return selfStringIntegerPart
} else {
if selfStringDecimalPart.count == numberOfDecimals {
return [selfStringIntegerPart,
selfStringDecimalPart].joined(separator: ".")
} else {
if selfStringDecimalPart.count > numberOfDecimals {
return [selfStringIntegerPart,
String(selfStringDecimalPart.prefix(numberOfDecimals))].joined(separator: ".")
} else {
let difference = numberOfDecimals - selfStringDecimalPart.count
let addedCharacters = [Character].init(repeating: "0", count: difference)
return [selfStringIntegerPart,
selfStringDecimalPart+addedCharacters].joined(separator: ".")
}
}
}
}
}
它可能看起来很老派，但我的所有测试都通过了：
func test_GivenADecimalNumber_ThenAssertNumberOfDecimalsWanted() {
//No decimals
XCTAssertEqual(Float(3).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Double(3).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 0), 3)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 0), 3)
//numberOfDecimals == totalNumberOfDecimals
XCTAssertEqual(Float(3.00).with(numberOfDecimals: 2), 3.00)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 2), 3.09)
XCTAssertEqual(Float(3.01).with(numberOfDecimals: 2), 3.01)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 3), 3.999)
XCTAssertEqual(Float(3.991).with(numberOfDecimals: 3), 3.991)
XCTAssertEqual(Double(3.00).with(numberOfDecimals: 2), 3.00)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 2), 3.09)
XCTAssertEqual(Double(3.01).with(numberOfDecimals: 2), 3.01)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 3), 3.999)
XCTAssertEqual(Double(3.991).with(numberOfDecimals: 3), 3.991)
//numberOfDecimals < totalNumberOfDecimals
XCTAssertEqual(Float(3.00).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Float(3.01).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 2), 3.99)
XCTAssertEqual(Float(3.991).with(numberOfDecimals: 2), 3.99)
XCTAssertEqual(Double(3.00).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Double(3.01).with(numberOfDecimals: 1), 3.0)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 2), 3.99)
XCTAssertEqual(Double(3.991).with(numberOfDecimals: 2), 3.99)
//numberOfDecimals > totalNumberOfDecimals
XCTAssertEqual(Float(3.00).with(numberOfDecimals: 3), 3.000)
XCTAssertEqual(Float(3.09).with(numberOfDecimals: 3), 3.090)
XCTAssertEqual(Float(3.01).with(numberOfDecimals: 3), 3.010)
XCTAssertEqual(Float(3.999).with(numberOfDecimals: 4), 3.9990)
XCTAssertEqual(Float(3.991).with(numberOfDecimals: 4), 3.9910)
XCTAssertEqual(Double(3.00).with(numberOfDecimals: 3), 3.000)
XCTAssertEqual(Double(3.09).with(numberOfDecimals: 3), 3.090)
XCTAssertEqual(Double(3.01).with(numberOfDecimals: 3), 3.010)
XCTAssertEqual(Double(3.999).with(numberOfDecimals: 4), 3.9990)
XCTAssertEqual(Double(3.991).with(numberOfDecimals: 4), 3.9910)
}
func test_GivenADecimal_ThenAssertStringValueWithDecimalsWanted() {
//No decimals
XCTAssertEqual(Float(3).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Double(3).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 0), "3")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 0), "3")
//numberOfDecimals == totalNumberOfDecimals
XCTAssertEqual(Float(3.00).string(numberOfDecimals: 2), "3.00")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 2), "3.09")
XCTAssertEqual(Float(3.01).string(numberOfDecimals: 2), "3.01")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 3), "3.999")
XCTAssertEqual(Float(3.991).string(numberOfDecimals: 3), "3.991")
XCTAssertEqual(Double(3.00).string(numberOfDecimals: 2), "3.00")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 2), "3.09")
XCTAssertEqual(Double(3.01).string(numberOfDecimals: 2), "3.01")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 3), "3.999")
XCTAssertEqual(Double(3.991).string(numberOfDecimals: 3), "3.991")
//numberOfDecimals < totalNumberOfDecimals
XCTAssertEqual(Float(3.00).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Float(3.01).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 2), "3.99")
XCTAssertEqual(Float(3.991).string(numberOfDecimals: 2), "3.99")
XCTAssertEqual(Double(3.00).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Double(3.01).string(numberOfDecimals: 1), "3.0")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 2), "3.99")
XCTAssertEqual(Double(3.991).string(numberOfDecimals: 2), "3.99")
//numberOfDecimals > totalNumberOfDecimals
XCTAssertEqual(Float(3.00).string(numberOfDecimals: 3), "3.000")
XCTAssertEqual(Float(3.09).string(numberOfDecimals: 3), "3.090")
XCTAssertEqual(Float(3.01).string(numberOfDecimals: 3), "3.010")
XCTAssertEqual(Float(3.999).string(numberOfDecimals: 4), "3.9990")
XCTAssertEqual(Float(3.991).string(numberOfDecimals: 4), "3.9910")
XCTAssertEqual(Double(3.00).string(numberOfDecimals: 3), "3.000")
XCTAssertEqual(Double(3.09).string(numberOfDecimals: 3), "3.090")
XCTAssertEqual(Double(3.01).string(numberOfDecimals: 3), "3.010")
XCTAssertEqual(Double(3.999).string(numberOfDecimals: 4), "3.9990")
XCTAssertEqual(Double(3.991).string(numberOfDecimals: 4), "3.9910")
}
方式一：如果你不想为此创建任何新函数，你可以通过这种方式直接获取舍入值。
var roundedValue = (decimalValue * pow(10.0, Double(numberOfPlaces))).rounded())/pow(10.0, Double(numberOfPlaces)
示例：
var numberOfPlaces = 2
var decimalValue = 13312.2423423523523
print("\(((decimalValue * pow(10.0, Double(numberOfPlaces))).rounded())/pow(10.0, Double(numberOfPlaces)))")
结果：13312.24
方式二：如果只想打印，可以使用：
print(String(format: "%.\(numberOfPlaces)f",decimalValue))
例子
var numberOfPlaces = 4
var decimalValue = 13312.2423423523523
print(String(format: "%.\(numberOfPlaces)f",decimalValue))