typescript - 如何使用泛型缩小 TypeScript 联合类型

Question

我有一个更新状态的泛型函数(用例是动态处理 React 中表的更新)，并使用泛型来确保调用该函数是类型安全的，但我不明白为什么 TypeScript 无法编译。
在函数本身中，Typescript 似乎没有使用所有可用信息来缩小类型，而是认为我仍在使用完整的联合。
它清楚地知道足够的参数来判断函数是否被正确调用，那么为什么在对实际实现进行类型检查时会失败呢？
在 row[field] = value 处失败.
最小示例:

type First = { a: string; b: string; c: string }
type Second = { b: string; c: string; d: string }
type Third = { c: string; d: string; e: string }

type State = {
  first: First[]
  second: Second[]
  third: Third[]
}

const update = <
  S extends State,
  TK extends keyof State,
  T extends S[TK],
  R extends T[number],
  F extends keyof R
>(
  state: State,
  tagName: TK,
  rowIndex: number,
  field: F,
  value: R[F]
) => {
  // fine
  const tag = state[tagName]

  // fine
  const row = tag[rowIndex]
  // keyof typeof row is now 'c'

  // TYPE ERROR
  row[field] = value
}

const state: State = {
  first: [{ a: "", b: "", c: "" }],
  second: [{ b: "", c: "", d: "" }],
  third: [{ c: "", d: "", e: "" }],
}

// this succeeds as expected
update(state, "first", 0, "a", "new")

// and this fails as expected
// @ts-expect-error
update(state, "second", 0, "a", "new")

Playground

Answer 1

尝试这个:

const update = <
  TK extends keyof State,
  F extends keyof State[TK][number]
>(
  state: State,
  tagName: TK,
  rowIndex: number,
  field: F,
  value: State[TK][number][F]
) => {
  // fine
  const tag = state[tagName]

  // fine
  // The type annotation here is required
  // (otherwise the type is inferred as First | Second | Third)
  const row: State[TK][number] = tag[rowIndex]

  // now also fine
  row[field] = value
}