kin*_*olo 185 class switch-statement swift
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.
TIA,彼得.
Rob*_*ier 393
你绝对可以is在一个switch块中使用.请参阅Swift编程语言中的"为任何和AnyObject类型转换"(当然,它不限于此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")
}
}
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Abh*_*eet 44
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)")
}
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Prc*_*ela 26
如果您没有值,只需要任何类:
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
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更新swift 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
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Dan*_*iel 11
我喜欢这种语法:
switch thing {
case _ as Int: print("thing is Int")
case _ as Double: print("thing is Double")
}
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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")
}
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