The protocol to which all classes implicitly conform.

You use AnyObject when you need the flexibility of an untyped object or when you use bridged Objective-C methods and properties that return an untyped result. AnyObject can be used as the concrete type for an instance of any class, class type, or class-only protocol. For example:

class FloatRef {
    let value: Float
    init(_ value: Float) {
        self.value = value
    }
}

let x = FloatRef(2.3)
let y: AnyObject = x
let z: AnyObject = FloatRef.self

AnyObject can also be used as the concrete type for an instance of a type that bridges to an Objective-C class. Many value types in Swift bridge to Objective-C counterparts, like String and Int.

let s: AnyObject = "This is a bridged string." as NSString
print(s is NSString)
// Prints "true"

let v: AnyObject = 100 as NSNumber
print(type(of: v))
// Prints "__NSCFNumber"

The flexible behavior of the AnyObject protocol is similar to Objective-C's id type. For this reason, imported Objective-C types frequently use AnyObject as the type for properties, method parameters, and return values.

Casting AnyObject Instances to a Known Type

Objects with a concrete type of AnyObject maintain a specific dynamic type and can be cast to that type using one of the type-cast operators (as, as?, or as!).

This example uses the conditional downcast operator (as?) to conditionally cast the s constant declared above to an instance of Swift's String type.

if let message = s as? String {
    print("Succesful cast to String: \(message)")
}
// Prints "Succesful cast to String: This is a bridged string."

If you have prior knowledge that an AnyObject instance has a particular type, you can use the unconditional downcast operator (as!). Performing an invalid cast triggers a runtime error.

let message = s as! String
print("Succesful cast to String: \(message)")
// Prints "Succesful cast to String: This is a bridged string."

let badCase = v as! String
// Runtime error

Casting is always safe in the context of a switch statement.

let mixedArray: [AnyObject] = [s, v]
for object in mixedArray {
    switch object {
    case let x as String:
        print("'\(x)' is a String")
    default:
        print("'\(object)' is not a String")
    }
}
// Prints "'This is a bridged string.' is a String"
// Prints "'100' is not a String"

Accessing Objective-C Methods and Properties

When you use AnyObject as a concrete type, you have at your disposal every @objc method and property---that is, methods and properties imported from Objective-C or marked with the @objc attribute. Because Swift can't guarantee at compile time that these methods and properties are actually available on an AnyObject instance's underlying type, these @objc symbols are available as implicitly unwrapped optional methods and properties, respectively.

This example defines an IntegerRef type with an @objc method named getIntegerValue().

class IntegerRef {
    let value: Int
    init(_ value: Int) {
        self.value = value
    }

    @objc func getIntegerValue() -> Int {
        return value
    }
}

func getObject() -> AnyObject {
    return IntegerRef(100)
}

let obj: AnyObject = getObject()

In the example, obj has a static type of AnyObject and a dynamic type of IntegerRef. You can use optional chaining to call the @objc method getIntegerValue() on obj safely. If you're sure of the dynamic type of obj, you can call getIntegerValue() directly.

let possibleValue = obj.getIntegerValue?()
print(possibleValue)
// Prints "Optional(100)"

let certainValue = obj.getIntegerValue()
print(certainValue)
// Prints "100"

If the dynamic type of obj doesn't implement a getIntegerValue() method, the system returns a runtime error when you initialize certainValue.

Alternatively, if you need to test whether obj.getIntegerValue() exists, use optional binding before calling the method.

if let f = obj.getIntegerValue {
    print("The value of 'obj' is \(f())")
} else {
    print("'obj' does not have a 'getIntegerValue()' method")
}
// Prints "The value of 'obj' is 100"

See Also: AnyClass