Question

这是我的代码：

var animalArray = ["cow","pig"]

switch animalArray {
case ["cow","pig"],["pig","cow"]:
    println("You Win!")
default:
    println("Keep Trying")

我收到错误：＆＃34;输入＆＃39;数组＆＃39;不符合协议＆＃39; IntervalType＆＃39;＆＃34;对于线＆＃34;案例[＆＃34;牛＆＃34;，＆＃34;猪＆＃34;]，[＆＃34;猪＆＃34;，＆＃34;牛＆＃34;]：＆＃ 34 ;.我究竟做错了什么？

Answer 1

你不能用数组做到这一点。但您可以使用contains()方法进行检查并迭代要测试的数组（此处为secondArray）：

var animalArray:[String] = ["cow", "pig"]
var secondArray:[String] = ["cow", "test"]

for s in secondArray{
    if(contains(animalArray, s)){
        println("animalArray Contains \(s)")
    }
}

Answer 2

switch语句需要Int。想一想：

var animalDict: [String: Int] = ["cow": 0,"pig": 1]
var animalSelection: Int = animalDict["cow"]!

switch animalSelection {
case 0:
    println("The Cow Wins!")
case 1:
    println("The Pig Wins!")
default:
    println("Keep Trying")
}

//prints "The Cow Wins!"

编辑1：

感谢大家的评论。我认为这是更健壮的代码：

var animalDict: [String: Int] = ["cow": 0,"pig": 1]
var animalSelection: Int? = animalDict["horse"]

if animalSelection as Int? != nil {
   switch animalSelection! {
   case 0:
       println("The Cow Wins!")
   case 1:
       println("The Pig Wins!")
   default:
       println("Keep Trying")
   }
} else {
    println("Keep Trying")
}

//prints "Keep Trying"

如果我说：

，它仍会打印The Cow Wins

var animalSelection:Int? = animalDict["cow"]

编辑2：

根据@ AirSpeedVelocity的评论，我测试了以下代码。比我自己的代码更优雅：

var animalDict: [String: Int] = ["cow": 0,"pig": 1]
var animalSelection = animalDict["horse"]

switch animalSelection {
case .Some(0):
    println("The Cow Wins!")
case .Some(1):
    println("The Pig Wins!")
case .None:
    println("Not a valid Selection")
default:
    println("Keep Trying")
}

Answer 3

如果您想要的是比较两个数组而不管其条目的顺序如何，那么我建议如下：

      var referenceAnimal = ["cow", "pig"]
      var animalsToTest = ["pig", "cow"]


      sort(&referenceAnimal)
      sort(&animalsToTest)


      if referenceAnimal == animalsToTest {
          println("You Win!")
      } else {
          println("Keep Trying")
      }

Answer 4

如果订单不重要，则可以这样做。

import Foundation


let animalArray = ["cow","pig"]

extension Array where Element == String  {
  static func ~=(pattern: Array, value: Array) -> Bool {
    return pattern == value
  }
}

switch animalArray {
case ["cow", "pig"], ["pig","cow"]:
    print("You Win!")
default:
    print("Keep Trying")
}

请记住，这会将它应用于您可能希望将其包装的所有Array<String>上，或者将其应用于所有Equatable。

Answer 5

您可以通过适当地重载switch来使~=能够匹配数组：

func ~=<T: Equatable>(lhs: [T], rhs: [T]) -> Bool {
    return lhs == rhs
}

var animalArray = ["cow","pig"]

switch animalArray {
case ["cow","pig"],["pig","cow"]:
    println("You Win!”)  // this will now match
default:
    println("Keep Trying")
}

这是否是一个好主意是值得怀疑的。

Answer 6

这与@Christian Woerz的答案几乎完全相同，但在这些情况下我是reduce的傻逼。

var animalArray = ["cow","pig"]
var answerArray = ["pig","cow"]
let isCorrect = answerArray.reduce(true) { bool, animal in
    return bool && contains(animalArray, animal)
}

Answer 7

为了将来参考，将开关包装在for-in循环中是另一种选择。

var animalArray = ["cow", "pig"]

for animal in animalArray {
    switch animal {
    case "cow":
        print("win")
    case "pig":
        print("lose")
    default:
        print("try again")
    }
}

如何为数组进行切换？

7 个答案:

编辑1：

编辑2：