any를 사용한 존재 형과 형 삭제

존재 타입이란 무엇인가요?

protocol Drawable { func draw() }
struct Line: Drawable { func draw() { print("Line") } }
let d: any Drawable = Line()
d.draw()

any 키워드

func render(_ shape: any Drawable) {
  shape.draw()
}
render(Line())

존재 타입의 비용

// Existential – dynamic dispatch, potential heap alloc
var shapes: [any Drawable] = [Line(), Line()]
// Generic – static dispatch, no boxing
func drawGeneric<T: Drawable>(_ s: T) { s.draw() }

PAT와 존재 타입의 문제

protocol Container { associatedtype Item }
// var c: any Container // Swift 5.7 allows with "any" but loses Item info

타입 소거 패턴

struct AnyDrawable: Drawable {
  private let _draw: () -> Void
  init<D: Drawable>(_ d: D) { _draw = d.draw }
  func draw() { _draw() }
}

AnySequence와 AnyPublisher

import Combine
let subject = PassthroughSubject<Int, Never>()
let pub: AnyPublisher<Int, Never> = subject.eraseToAnyPublisher()

eraseToAnyPublisher()

func fetch() -> AnyPublisher<String, Error> {
  URLSession.shared.dataTaskPublisher(for: URL(string: "https://example.com")!)
    .map { String(data: $0.data, encoding: .utf8) ?? "" }
    .eraseToAnyPublisher()
}

존재 타입이 적합한 경우

var widgets: [any Drawable] = [] // mixed concrete types OK

존재 타입보다 제네릭 우선하기

// Prefer this:
func draw<T: Drawable>(_ item: T) { item.draw() }
// Over this:
func draw(_ item: any Drawable) { item.draw() }

존재 타입 열기

func drawGeneric<T: Drawable>(_ d: T) { d.draw() }
let d: any Drawable = Line()
drawGeneric(d) // compiler opens the existential automatically

타입 소거와 불투명 타입 비교

let erased: any Drawable = Line()   // runtime type unknown
let opaque: some Drawable = Line()   // compile-time fixed type

빠른 확인

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