符合协议时类型别名声明的冗余重复（第 2 部分）

Question

protocol Destinationable: Hashable {
    associatedtype D: Hashable
    var destination: D { get }
}

protocol Graph {
    associatedtype Edge: Destinationable
    subscript(node: D) -> Set<Edge>! { get set }
}

extension Graph {
    typealias D = Edge.D
}

struct UndirectedGraph<Edge: Destinationable>: Graph {
    typealias D = Edge.D // Why should we again declare this typealias!?
    
    private var storage: [D: Set<Edge>]
    
    subscript(node: D) -> Set<Edge>! {
        get { storage[node] }
        set { storage[node] = newValue }
    }
}

这行得通。但是如果我在结构中删除 typealias D = Edge.D 的重新声明，我会得到一个与下标相关的编译错误：

Unsupported recursion for reference to type alias 'D' of type 'UndirectedGraph'

为什么会这样？什么递归？？？

Answer 1

我不确定什么是递归，但你应该避免在 protocol extensions 中声明类型别名，而是使用相同类型的约束：

protocol Destinationable: Hashable {
    associatedtype D: Hashable
    var destination: D { get }
}

protocol Graph {
    associatedtype D
    associatedtype Edge: Destinationable where D == Edge.D
    subscript(node: D) -> Set<Edge>! { get set }
}

struct UndirectedGraph<Edge: Destinationable>: Graph {
    private var storage: [D: Set<Edge>]
    
    subscript(node: Edge.D) -> Set<Edge>! {
        get { storage[node] }
        set { storage[node] = newValue }
    }
}

编译得很好，但在你的情况下：

protocol Destinationable: Hashable {
    associatedtype D: Hashable
    var destination: D { get }
}

protocol Graph {
    associatedtype Edge: Destinationable
    subscript(node: Edge.D) -> Set<Edge>! { get set }
}

struct UndirectedGraph<Edge: Destinationable>: Graph {
    private var storage: [Edge.D: Set<Edge>]
    
    subscript(node: Edge.D) -> Set<Edge>! {
        get { storage[node] }
        set { storage[node] = newValue }
    }
}

可能就够了。

符合协议时类型别名声明的冗余重复（第 2 部分）

Redundant duplication of typealiase declarations when conforms to a protocol (Part 2)

protocols

type-alias

swift