在 Swift 中匹配对象的数据类型

Question

如何在 Swift 中匹配对象的数据类型？

喜欢：

var xyz : Any
    xyz = 1;
    switch xyz
 {
    case let x where xyz as?AnyObject[]:
        println("\(x) is AnyObject Type")
    case let x where xyz as?String[]:
        println("\(x) is String Type")
    case let x where xyz as?Int[]:
        println("\(x) is Int Type")
    case let x where xyz as?Double[]:
        println("\(x) is Double Type")
    case let x where xyz as?Float[]:
        println("\(x) is Float Type")
    default:println("None")
    }

在这种情况下切换案例运行默认案例

Answer 1

将 var xyz : AnyObject 更改为 var xyz : Any 并添加它将匹配此案例

case let x as Int:

---

来自 REPL

  1> var a : Any = 1
a: Int = <read memory from 0x7fec8ad8bed0 failed (0 of 8 bytes read)>
  2> switch a { case let x as Int: println("int"); default: println("default"); }
int

---

来自The Swift Programming Language

您可以在 switch 语句的 case 中使用 is 和 as 运算符来 发现已知的常量或变量的特定类型 只能是 Any 或 AnyObject 类型。下面的示例迭代 things 数组中的项目，并使用 a 查询每个项目的类型 切换语句。一些 switch 语句的 case 绑定了它们的 将值匹配到指定类型的常量以启用其值 待打印：

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)")
    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")
    }
}

// zero as an Int
// zero as a Double
// an integer value of 42
// a positive double value of 3.14159
// a string value of "hello"
// an (x, y) point at 3.0, 5.0
// a movie called 'Ghostbusters', dir. Ivan Reitman

---

注意：

var xyz : AnyObject = 1

会给你NSNumber 因为Int 不是对象，所以它会自动将它转换为NSNumber 这是对象

Answer 2

提出一个有趣的“case is”用法，即“case is Int, is String”，"," 行为类似于 OR 运算符。

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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