swift4:

if obj is MyClass{
    // then object type is MyClass Type
}

我有两种方法:

if let thisShape = aShape as? Square 

Or:

aShape.isKindOfClass(Square)

下面是一个详细的例子:

class Shape { }
class Square: Shape { } 
class Circle: Shape { }

var aShape = Shape()
aShape = Square()

if let thisShape = aShape as? Square {
    println("Its a square")
} else {
    println("Its not a square")
}

if aShape.isKindOfClass(Square) {
    println("Its a square")
} else {
    println("Its not a square")
}

编辑:3现在:

let myShape = Shape()
if myShape is Shape {
    print("yes it is")
}

如果你有这样的响应:

{
  "registeration_method": "email",
  "is_stucked": true,
  "individual": {
    "id": 24099,
    "first_name": "ahmad",
    "last_name": "zozoz",
    "email": null,
    "mobile_number": null,
    "confirmed": false,
    "avatar": "http://abc-abc-xyz.amazonaws.com/images/placeholder-profile.png",
    "doctor_request_status": 0
  },
  "max_number_of_confirmation_trials": 4,
  "max_number_of_invalid_confirmation_trials": 12
}

你想检查值is_stuck将被读取为AnyObject，你所要做的就是这个

if let isStucked = response["is_stucked"] as? Bool{
  if isStucked{
      print("is Stucked")
  }
  else{
      print("Not Stucked")
 }
}

在Swift 2.2 - 5你现在可以做:

if object is String
{
}

然后过滤你的数组:

let filteredArray = originalArray.filter({ $0 is Array })

如果你有多个类型需要检查:

    switch object
    {
    case is String:
        ...

    case is OtherClass:
        ...

    default:
        ...
    }

let originalArray : [Any?] = ["Hello", "World", 111, 2, nil, 3.34]
let strings = originalArray.compactMap({ $0 as? String })

print(strings)
//printed: ["Hello", "World"]

为什么不用这样的东西呢

fileprivate enum types {
    case typeString
    case typeInt
    case typeDouble
    case typeUnknown
}

fileprivate func typeOfAny(variable: Any) -> types {
    if variable is String {return types.typeString}
    if variable is Int {return types.typeInt}
    if variable is Double {return types.typeDouble}
    return types.typeUnknown
}

在Swift 3中。