How are Int and String accepted as AnyHashable?

Question

How come I can do this?

    var dict = [AnyHashable : Int]()
    dict[NSObject()] = 1
    dict[\"\"] = 2

This implies that NSObject

Answer 1

You can find this code, when you cmd-click on [ or ] of dict[NSObject()] = 1 in the Swift editor of Xcode (8.2.1, a little different in 8.3 beta):

extension Dictionary where Key : _AnyHashableProtocol {

    public subscript(key: _Hashable) -> Value?

    public mutating func updateValue(_ value: Value, forKey key: ConcreteKey) -> Value?

    public mutating func removeValue(forKey key: ConcreteKey) -> Value?
}

_AnyHashableProtocol and _Hashable are hidden types, so you may need to check the Swift source code. Simply:

• _Hashable is a hidden super-protocol of Hashable, so, Int, String, NSObject or all other Hashable types conform to _Hashable.

• _AnyHashableProtocol is a hidden protocol, where AnyHashable is the only type which conforms to _AnyHashableProtocol.

So, the extension above is very similar to this code in Swift 3.1.

extension Dictionary where Key == AnyHashable {

    //...
}

You see, when you write such code like this:

    var dict = [AnyHashable : Int]()
    dict[NSObject()] = 1
    dict[""] = 2

You are using the subscript operator defined in the extension.