CODEWITHSUNDEEP
arraysswiftobject
rghome
rghome

Reputation: 8819

Why is a Swift Array<Int> compatible with AnyObject?

Consider the code:

struct S {
    var f : Int64 = 0
}

...

let coder : NSCoder = someCoder ...
let a : [Int] = []
coder.encodeObject(a)  // compiles
let b : [Int64] = []
coder.encodeObject(b)  // doesn't compile: not AnyObject
let s : [S] = []
coder.encodeObject(s)  // doesn't compile: not AnyObject

Note that Int is defined as a struct.

So [Int] is object, but [Int64] is not and neither is my array of simple structures.

What is special about Int?

Upvotes: 2

Views: 289

Answers (3)

matt
matt

Reputation: 534950

Int is bridged (as are UInt and Float and Double, as well as Bool). This means that they are wrapped up in an NSNumber for you, automatically, when you cast to AnyObject, and vice versa.

Other numeric types, alas, are not.

In turn, you also get to take advantage of array-casting syntactic sugar. An NSArray must consist of Objective-C objects, such as NSNumber. When you start with a Swift array, instead of having to wrap the elements up in an NSNumber yourself, as you do with an array of Int64, they are wrapped for you when you cast/bridge a Swift array to an NSArray.

Upvotes: 3

vadian
vadian

Reputation: 285069

If Foundation framework is imported, Int (unlike Int64) is implicitly bridged to NSNumber which conforms to AnyObject

Upvotes: 1

Rob Napier
Rob Napier

Reputation: 299275

If you import Foundation (which you must be, because you reference NSCoder) then [Int] is implicitly bridged to NSArray because Int is implicitly bridged to NSNumber. Int64 and your non-objc structs are not implicitly bridged to ObjC types, so arrays of those are not bridged to NSArray.

Upvotes: 5

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