Swift：在switch语句中测试类类型

Question

在Swift中，您可以使用“is”检查对象的类类型。如何将其合并到“开关”块中？

我认为这是不可能的，所以我想知道最好的方法是什么。

Answer 1

你绝对可以在is区块中使用switch。请参阅Swift编程语言中的“为任意和AnyObject类型转换”（当然，它不仅限于Any）。他们有一个广泛的例子：

for thing in things {
    switch thing {
    case 0 as Int:
        println("zero as an Int")
    case 0 as Double:
        println("zero as a Double")
    case let someInt as Int:
        println("an integer value of \(someInt)")
    case let someDouble as Double where someDouble > 0:
        println("a positive double value of \(someDouble)")
// here it comes:
    case is Double:
        println("some other double value that I don't want to print")
    case let someString as String:
        println("a string value of \"\(someString)\"")
    case let (x, y) as (Double, Double):
        println("an (x, y) point at \(x), \(y)")
    case let movie as Movie:
        println("a movie called '\(movie.name)', dir. \(movie.director)")
    default:
        println("something else")
    }
}

Answer 2

提出“case is - case is Int，is String：”操作的示例，其中多个案例可以一起使用，以对类似对象类型执行相同的活动。这里“，”将类型分开，就像OR运算符一样。

switch value{
case is Int, is String:
    if value is Int{
        print("Integer::\(value)")
    }else{
        print("String::\(value)")
    }
default:
    print("\(value)")
}

Demo Link

Answer 3

如果您没有值，只需要任何对象：

迅速4

func test(_ val:Any) {
    switch val {
    case is NSString:
        print("it is NSString")
    case is String:
        print("it is a String")
    case is Int:
        print("it is Int")
    default:
        print(val)
    }
}


let str: NSString = "some nsstring value"
let i:Int=1
test(str) // it is NSString
test(i) // it is Int

Answer 4

我喜欢这种语法：

switch thing {
case _ as Int: print("thing is Int")
case _ as Double: print("thing is Double")
}

因为它使您可以快速扩展功能，如下所示：

switch thing {
case let myInt as Int: print("\(myInt) is Int")
case _ as Double: print("thing is Double")
}