如何为数组进行切换？

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："Type 'Array' does not conform to protocol 'IntervalType'" 行 "case ["cow","pig"],["pig","cow"]:"。我究竟做错了什么？

Answer 1

你不能用数组来做到这一点。但是您可以使用 contains() 方法检查它并遍历要测试的数组（此处 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。想想这个：

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：

，它仍然会打印 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。比我自己的代码优雅多了：

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

如果您想比较两个数组而不考虑它们的条目顺序，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

您可以通过适当重载 ~= 使 switch 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")
}

尽管这是否是个好主意值得怀疑。

Answer 5

这与@Christian Woerz 的回答几乎相同，但在这些情况下我是 reduce 的傻瓜。

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

Answer 6

如果顺序不重要，您可以这样做。

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，这会将它应用到所有 Array<String>，您可能想将它们包装起来，或者换一种方式将它应用到所有 Equatable。

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？

How to make a switch for an array?

arrays

switch-statement

swift

编辑 1：

编辑 2：