Using Optional, your enum type goes away. (I think it is a great solution
unless you need something more than .element and .none in real life.) Great to
get all that optional machinery for missing values for free! Then you can
constrain elements simply from the Element protocol as in as in:
protocol Element {
func compute() -> Int
}
struct ElementType1: Element {
func compute() -> Int {
return 1
}
}
struct ElementType2: Element {
func compute() -> Int {
return 2
}
}
var childElements: [Element?] = []
childElements.append(ElementType1())
childElements.append(ElementType2())
childElements.append(nil)
childElements.flatMap { $0 }.forEach { print($0.compute()) }
1
2
> On Dec 28, 2016, at 4:10 PM, Brandon Knope via swift-users
> <[email protected]> wrote:
>
> Aren’t I losing the ability to enforce what is going into this enum’s
> associated value then?
>
> Brandon
>
>> On Dec 28, 2016, at 7:05 PM, Nevin Brackett-Rozinsky
>> <[email protected] <mailto:[email protected]>>
>> wrote:
>>
>> It will work if you change the enum declaration to:
>>
>> enum ElementNode<T>
>>
>> In other words, let the enum hold arbitrary unconstrained associated types,
>> and then make your APIs utilize instances of the enum with the associated
>> type constrained to a protocol.
>>
>> The specific example you provide is essentially equivalent to:
>>
>> var childElements = [Element?]()
>>
>> Nevin
>>
>>
>> On Wed, Dec 28, 2016 at 6:41 PM, Brandon Knope via swift-users
>> <[email protected] <mailto:[email protected]>> wrote:
>> I don’t understand why this is a problem
>>
>> protocol Element {
>>
>> }
>>
>> enum ElementNode<T: Element> {
>> case element(T)
>> case empty
>> }
>>
>> var childElements = [ElementNode<Element>]()
>>
>> I need to represent an array of my nodes that could be multiple kinds of
>> elements
>>
>> Is there a workaround?
>>
>> Brandon
>>
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>>
>>
>
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