Swift：在 switch 语句中测试类类型

发布于09月08日

在Swift中，可以使用"is"判断对象的类类型.如何将其合并到"switch "块中？

我认为这是不可能的，所以我想知道最好的解决办法是什么.

推荐答案

你绝对可以在switch街区使用is.请参阅Swift编程语言中的"任意和任意对象的类型转换"(当然不限于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")
    }
}