我不知道小数点后两位，但这里是你如何打印小数位为零的浮点数，所以我可以想象它可以是2位，3位……(注意:你必须将CGFloat转换为Double来传递给String(format:)否则它会看到一个0的值)

func logRect(r: CGRect, _ title: String = "") {
    println(String(format: "[ (%.0f, %.0f), (%.0f, %.0f) ] %@",
        Double(r.origin.x), Double(r.origin.y), Double(r.size.width), Double(r.size.height), title))
}

这里有一个“纯粹的”快速解决方案

 var d = 1.234567
operator infix ~> {}
@infix func ~> (left: Double, right: Int) -> String {
    if right == 0 {
        return "\(Int(left))"
    }
    var k = 1.0
    for i in 1..right+1 {
        k = 10.0 * k
    }
    let n = Double(Int(left*k)) / Double(k)
    return "\(n)"
}
println("\(d~>2)")
println("\(d~>1)")
println("\(d~>0)")

@Christian Dietrich):

而不是:

var k = 1.0
    for i in 1...right+1 {
        k = 10.0 * k
    }
let n = Double(Int(left*k)) / Double(k)
return "\(n)"

也可以是:

let k = pow(10.0, Double(right))
let n = Double(Int(left*k)) / k
return "\(n)"

(更正:) 抱歉混淆* -当然这适用于双打。我认为，最实用的(如果你想让数字四舍五入，而不是被切断)应该是这样的:

infix operator ~> {}
func ~> (left: Double, right: Int) -> Double {
    if right <= 0 {
        return round(left)
    }
    let k = pow(10.0, Double(right))
    return round(left*k) / k
}

仅对于Float，只需将Double替换为Float, pow替换为powf, round替换为roundf。 更新:我发现它是最实用的使用返回类型Double而不是字符串。它的工作原理与字符串输出相同，即:

println("Pi is roughly \(3.1415926 ~> 3)")

印花:圆周率大约是3.142 所以你可以用同样的方式使用它的字符串(你甚至可以写:println(d ~> 2))，但另外你也可以用它直接舍入值，即:

d = Double(slider.value) ~> 2

或者任何你需要的…

少打字方式:

func fprint(format: String, _ args: CVarArgType...) {
    print(NSString(format: format, arguments: getVaList(args)))
}

那么Double和CGFloat类型的扩展呢?

extension Double {

   func formatted(_ decimalPlaces: Int?) -> String {
      let theDecimalPlaces : Int
      if decimalPlaces != nil {
         theDecimalPlaces = decimalPlaces!
      }
      else {
         theDecimalPlaces = 2
      }
      let theNumberFormatter = NumberFormatter()
      theNumberFormatter.formatterBehavior = .behavior10_4
      theNumberFormatter.minimumIntegerDigits = 1
      theNumberFormatter.minimumFractionDigits = 1
      theNumberFormatter.maximumFractionDigits = theDecimalPlaces
      theNumberFormatter.usesGroupingSeparator = true
      theNumberFormatter.groupingSeparator = " "
      theNumberFormatter.groupingSize = 3

      if let theResult = theNumberFormatter.string(from: NSNumber(value:self)) {
         return theResult
      }
      else {
         return "\(self)"
      }
   }
}

用法:

let aNumber: Double = 112465848348508.458758344
Swift.print("The number: \(aNumber.formatted(2))")

打印:112 465 848 348 508.46

我目前为止最好的解决方案，以下是David的回答:

import Foundation

extension Int {
    func format(f: String) -> String {
        return String(format: "%\(f)d", self)
    }
}

extension Double {
    func format(f: String) -> String {
        return String(format: "%\(f)f", self)
    }
}

let someInt = 4, someIntFormat = "03"
println("The integer number \(someInt) formatted with \"\(someIntFormat)\" looks like \(someInt.format(someIntFormat))")
// The integer number 4 formatted with "03" looks like 004

let someDouble = 3.14159265359, someDoubleFormat = ".3"
println("The floating point number \(someDouble) formatted with \"\(someDoubleFormat)\" looks like \(someDouble.format(someDoubleFormat))")
// The floating point number 3.14159265359 formatted with ".3" looks like 3.142

我认为这是最类似swift的解决方案，将格式化操作直接绑定到数据类型上。很可能在某个地方有一个内置的格式化操作库，或者它很快就会发布。请记住，该语言仍处于测试阶段。