Question

我正在研究ts中的代数数据类型，并坚持输入匹配函数。问题的简化版本如下所示：

let rec = { a: 'a', b: 1 };

// just captures type of 'rec' and then reuses the type for eval
const makeEval = <Record>(rec: Record) => <
  Res,
  K extends keyof Record = keyof Record
>(
  val: Record,
  cases:
    | Cases<Record, Res>
    | (Cases<Record, Res, K> & { else: (r: Record) => Res })
): Res => (undefined as any) as Res;

export type Cases<Record, Res, K extends keyof Record = keyof Record> = {
  [T in K]: (value: Record[T]) => Res
};

const evalMyRecord = makeEval(rec);

// this is fine
const a = evalMyRecord(rec, { a: s => s, b: n => n.toString() });

// err
const b = evalMyRecord(rec, { a: s => s, else: _ => 'why err?' });
// Property 'b' is missing in type '{ a: (s: string) => string; else: (_: { a: string; b: number; }) => string; }'.

// requires explicit subset
const b_ = evalMyRecord<string, 'a'>(rec, {
  a: s => s,
  else: _ => 'but explicit a is fine'
});

// full set is fine
const c = evalMyRecord(rec, {
  a: s => s,
  b: n => 'n',
  else: _ => 'fine too'
});

所以我想表达一个类型，它具有Record的所有键或其任何子集+ {else} case。

我知道Partial有一个解决方案：

type EvalCases<Record, Res> =
  | Cases<Record, Res>
  | (Partial<Cases<Record, Res>> & { else: (r: Record) => Res });

const b = evalMyRecord(rec, { a: s => s, else: _ => 'works' });

const b2 = evalMyRecord(rec, {
  a: s => s,
  b: undefined,
  else: _ => 'also works but weird'
});

但是：{b：undefined}看起来有点偏。如果任何情况不是一个正确的函数（如果可能的话），我想要编译器错误。

有关打字稿魔术的任何建议吗？

注意：概念上与Typescript cannot infer correct argument types for an object of functions关闭（但不相同）。

Answer 1

以下解决方案允许您拥有Cases<Record, Res>或Cases<Record, Res>的任何子集以及else属性。

type EnforcingPartial<T> = {
    [key in keyof T]: { [subKey in key]: T[key]; }
}[keyof T];

type EvalCases<Record, Res> = Cases<Record, Res> |
    (EnforcingPartial<Cases<Record, Res>> & { else: (r: Record) => Res });

const b = evalMyRecord(rec, { a: s => s, else: _ => 'works' });

const b2 = evalMyRecord(rec, {
    a: s => s,
    b: undefined, // ERROR HERE, b is incompatible
    else: _ => 'also works but weird'
});

typescript无法从使用中推断出对象的键的子集

1 个答案: