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typescript - 通过递归对象结构生成表示路径联合的元组类型

问题描述

下面是几个模板化的“业务”类型，分别用于表示复合状态和原子状态。

interface CompoundState<TName extends string, TChildren extends { [key: string]: AnyCompoundState | AnyAtomicState }> {
  type: 'parent'
  name: TName,
  children: TChildren,
};

type AnyCompoundState = CompoundState<string, { [key: string]: AnyCompoundState | AnyAtomicState }>;

interface AtomicState<TName extends string> {
  type: 'child',
  name: TName,
}

type AnyAtomicState = AtomicState<string>;

在我的应用程序中，这些类型将被组合以产生复合和原子状态的树状结构。以下是此类类型的示例：

type MyStateChart = CompoundState<'cs0', {
  cs1: CompoundState<'cs1', {
    as1: AtomicState<'as1'>,
    as2: AtomicState<'as2'>,
  }>
}>;

我想要完成的是生成一个元组联合来表示 type 暗示的可能的“路径” MyStateChart。可能的路径是元组，例如：

  1. ['cs0']- a 的有效路径CompoundState可能会或可能不会遍历到子级。
  2. ['cs0', 'cs1']- 和上面一样，我们不需要遍历叶子节点。
  3. ['cs0', 'cs1', 'as1']- 全深度
  4. ['cs0', 'cs1', 'as2']- 全深度

在实现这一目标的（大部分）失败尝试中，我采取了两种方法：

方法一：

type PathA<TNode extends AnyCompoundState | AnyAtomicState> = TNode extends AnyCompoundState
  ? {
    [K in keyof TNode['children']]: [TNode['name']] | [TNode['name'], PathA<TNode['children'][K]>]
  }[keyof TNode['children']]
  : [TNode['name']]

// Produces a type represented by nested tuple unions. I have been unable to 'flatten' this into distinct, fully-realized tuples
type TestPathA = PathA<MyStateChart>;

这会产生一种非常接近我想要的类型但我无法“扁平化”的类型：

type TestPathA = ["cs0"] | ["cs0", ["cs1"] | ["cs1", ["l1"]] | ["cs1", ["l2"]]]

方法二：

type Cons<H, T extends unknown[]> = ((h: H, ...tail: T) => unknown) extends ((...args: infer U) => unknown) ? U : never;

// Approach B: Approach that I hoped would work but complains with:
type PathB<TNode extends AnyCompoundState | AnyAtomicState> = TNode extends AnyCompoundState
  ? {
    [K in keyof TNode['children']]: [TNode['name']] | Cons<TNode['name'], PathB<TNode['children'][K]>>
  }[keyof TNode['children']]
  : [TNode['name']]

type TestPathB = PathB<MyStateChart>;

这种方法似乎是无限的，TypeScript 编译器抱怨：

"Type instantiation is excessively deep and possibly infinite.(2589)"

我能达到我想要的吗？如果有怎么办？

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标签： typescript

解决方案

正如@jcalz 在他的评论中指出的那样，这个问题可以通过与这个问题的答案相同的方法来解决。

这是应用于上述问题的样子：

type Cons<H, T> = T extends readonly any[] ?
  ((h: H, ...t: T) => void) extends ((...r: infer R) => void) ? R : never
  : never;

type Prev = [never, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
  11, 12, 13, 14, 15, 16, 17, 18, 19, 20, ...0[]]

type Paths<T extends AnyAtomicState | AnyCompoundState, D extends number = 10> = [D] extends [never] ? never : T extends AnyCompoundState ?
  { [K in keyof T['children']]-?: [T['name']] | (Paths<T['children'][K], Prev[D]> extends infer P ?
    P extends [] ? never : Cons<T['name'], P> : never
  ) }[keyof T['children']]
  : [T['name']];

type TestC = Paths<MyStateChart>;

产生以下类型：

type TestC = ["cs0"] | ["cs0", "cs1"] | ["cs0", "cs1", "l1"] | ["cs0", "cs1", "l2"]

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