在Swift中,您可以使用'is‘检查对象的类类型。如何将其合并到“switch”块中?
我认为这是不可能的,所以我想知道解决这个问题的最好方法是什么。
发布于 2014-09-08 20:47:52
您绝对可以在switch
块中使用is
。请参阅Swift编程语言中的“Any和AnyObject的类型转换”(当然,它并不限于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")
}
}
发布于 2018-12-13 21:50:35
我喜欢这个语法:
switch thing {
case _ as Int: print("thing is Int")
case _ as Double: print("thing is Double")
}
因为它提供了快速扩展功能的可能性,如下所示:
switch thing {
case let myInt as Int: print("\(myInt) is Int")
case _ as Double: print("thing is Double")
}
https://stackoverflow.com/questions/25724527
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