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37e204a1e3
It simply defers it to Syntax.
57 lines
1.7 KiB
Swift
57 lines
1.7 KiB
Swift
/// The fixpoint of `Syntax`.
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///
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/// `Syntax` is a non-recursive type parameterized by the type of its child nodes. Instantiating it to `Fix` makes it into a recursive tree by “tying the knot”—each child node of `Syntax<Fix, A>` is represented by a `Fix` which in turn contains a `Syntax<Fix, A>`. So in the same way that the `fix` function allows one to tie a non-recursive function into a recursive one, `Fix` allows one to tie a non-recursive type into a recursive one. Unfortunately, due to Swift’s lack of higher-rank types, this cannot currently be abstracted over the type which is made recursive, and thus it is hard-coded to `Syntax<Fix, A>` rather than provided by a type parameter `F` applied to `Fix<F>`.
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public enum Fix<A>: CustomDebugStringConvertible, CustomDocConvertible {
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/// A recursive instantiation of `Syntax`, unrolling another iteration of the recursive type.
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indirect case In(Syntax<Fix, A>)
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public var out: Syntax<Fix, A> {
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switch self {
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case let .In(s):
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return s
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}
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}
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// MARK: CustomDebugStringConvertible
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public var debugDescription: String {
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return ".In(\(String(reflecting: out)))"
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}
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// MARK: CustomDocConvertible
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public var doc: Doc {
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return out.doc
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}
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}
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// MARK: - Equality
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extension Fix {
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public static func equals(param: (A, A) -> Bool)(_ left: Fix, _ right: Fix) -> Bool {
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return Syntax.equals(ifLeaf: param, ifRecur: equals(param))(left.out, right.out)
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}
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}
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public func == <A: Equatable> (left: Fix<A>, right: Fix<A>) -> Bool {
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return Fix.equals(==)(left, right)
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}
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// MARK: - Hashing
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extension Fix {
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public func hash(param: A -> Hash) -> Hash {
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return out.hash(ifLeaf: param, ifRecur: { $0.hash(param) })
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}
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}
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extension Fix where A: Hashable {
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public var hash: Hash {
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return hash(Hash.init)
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}
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}
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