怪异的Swift 4型系统

Question

Can someone explain the best best way to figure out Int types, since Swift 4 I have so many troubles with JSON because of this

cause this is so common situation when I trying to map same json object to core data entities and PONSOs

so,

id: Int = json["id"] as? Int // works
id: Int32 = json["id"] as? Int32 // doesn't work

as a result I'm unable to satisfy both requirements at the same time. Can I anyhow avoid this absolutely senseless int situation in Swift 4?

Answer 1

The reason this is confusing is because as? has two completely orthogonal meanings in Swift, depending on the context. When dealing purely with Swift types, as? is a dynamic cast that returns non- nil only if the type is literally an instance of the type on the right-side of the as? operator:

import Foundation

let dict: [String : Any] = ["Foo" : 3 as Int]

print(dict["Foo"] as? Int)
print(dict["Foo"] as? Int32)
print(dict["Foo"] as? Int64)

This returns Optional(3) for the first log, and nil for the other two, because the type is an Int and not an Int32 or an Int64 .

However, when the type of the item being cast is an Objective-C type, then as? is no longer a strict dynamic cast, but instead causes bridging behavior:

import Foundation

let dict: [String : Any] = ["Foo" : 3 as NSNumber]

print(dict["Foo"] as? Int)
print(dict["Foo"] as? Int32)
print(dict["Foo"] as? Int64)

This returns Optional(3) for all three logs, because this is no longer a dynamic cast—indeed, an NSNumber instance is not a member of any of the three types to which we tried to cast. Instead, as? causes Swift to bridge the Objective-C type, NSNumber , to an appropriate Swift type if it can. Since Swift has logic to bridge NSNumber to Int , Int32 , and Int64 (along with a host of other numeric types), we get Optional(3) for all three logs. However, if you try to cast to something like Decimal for which there's no NSNumber bridging logic, you still get nil .

An interesting side-effect of this is that as? does not follow the transitive property:

let foo: Int = 3
print(foo as? NSNumber as? Int64) // Optional(3)
print(foo as? Int64)              // nil

Anyway, if you cast your values to NSNumber first, you should then be able to cast from there to any of the numeric types that the Objective-C bridge supports, which is probably what was going on under the hood with your old Swift 3 code. Alternatively, if you actually know the type that the value is supposed to be, you can use one of the other integer types' initializers:

let dict: [String : Any] = ["Foo" : 3 as Int]

let foo = (dict["Foo"] as? Int).map { Int32($0) }