TypeScript没有推断出正确的类型

Question

对于这件作品：

Promise.resolve('kromid')
  .then(all(identity))
  .then(([a]) => a.splita);

TypeScript没有提及a.splita的任何内容。我期待它失败：

  

Property＆＃39; splita＆＃39;类型＆＃39;字符串＆＃39;上不存在。你的意思是   ＆＃39;分裂＆＃39;？

以下是代码的其余部分：

function all<T1, Param>(a1: Res<Param, T1>): (p: Param) => Promise<[T1]>;
function all<T1, T2, Param>(a1: Res<Param, T1>, a2: Res<Param, T2>): (p: Param) => Promise<[T1, T2]>;
function all<T1, T2, T3, Param>(a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>): (p: Param) => Promise<[T1, T2, T3]>;
function all<T1, T2, T3, T4, Param>(a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>): (p: Param) => Promise<[T1, T2, T3, T4]>;
function all<T1, T2, T3, T4, T5, Param>(a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>): (p: Param) => Promise<[T1, T2, T3, T4, T5]>;
function all<T1, T2, T3, T4, T5, T6, Param>(a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>, a6: Res<Param, T6>): (p: Param) => Promise<[T1, T2, T3, T4, T5, T6]>;
function all<T1, T2, T3, T4, T5, T6, T7, Param>(a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>, a6: Res<Param, T6>, a7: Res<Param, T7>): (p: Param) => Promise<[T1, T2, T3, T4, T5, T6, T7]>;
function all<T1, T2, T3, T4, T5, T6, T7, T8, Param>(a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>, a6: Res<Param, T6>, a7: Res<Param, T7>, a8: Res<Param, T8>): (p: Param) => Promise<[T1, T2, T3, T4, T5, T6, T7, T8]>;
function all<T1, T2, T3, T4, T5, T6, T7, T8, T9, Param>(a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>, a6: Res<Param, T6>, a7: Res<Param, T7>, a8: Res<Param, T8>, a9: Res<Param, T9>): (p: Param) => Promise<[T1, T2, T3, T4, T5, T6, T7, T8, T9]>;
function all<T1, T2, T3, T4, T5, T6, T7, T8, T9, T10, Param>(a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>, a6: Res<Param, T6>, a7: Res<Param, T7>, a8: Res<Param, T8>, a9: Res<Param, T9>, a10: Res<Param, T10>): (p: Param) => Promise<[T1, T2, T3, T4, T5, T6, T7, T8, T9, T10]>;
function all<Param>(...values: Res<Param, any>[]): (p: Param) => Promise<any[]>;
function all<Param>(...values: Res<Param, any>[]): (p: Param) => Promise<any[]> {
  return param => Promise.all(values.map(obj => obj.apply ? obj(param) : obj));
}
type Res<I, O> = ((i: I) => O | Promise<O>) | O | Promise<O>


function identity<T>(a: T): T {
  return a;
}

你能发现问题所在吗？

Answer 1

感谢您指出我没有得到的问题。我确实更多地使用了你的代码，这是我的发现。类型推断不适用于TypeScript中的函数组合：

function identity<T>(a: T): T { return a; }
function toString<T>(a: T): string { return JSON.stringify(a); }

Promise.resolve('kromid')
  .then(promiseValue => {
    const composedFn = all(identity, toString);
//        ^^^^^^^^^^
//        T type of `promiseValue` was not inferred properly,
//        and was replaces with `any`
    return composedFn(promiseValue);
  })
  .then(([identityResult, toStringResult]) => {

  })

请注意，类型推断针对函数(T) => T而破坏，但适用于非泛型返回类型函数(T) => string。在路上，identityResult的类型为any，而toStringResult的类型为string。

我认为，这个GitHub issue涵盖了真正的根本原因。对不起，我的回答不是很有帮助。类型推断对我来说是最复杂的话题，这就是我为什么要深入研究你的例子。

抓我的旧回答

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<击>

为什么会这样？

我认为，关键在于这一行：

  .then(all(identity))

相当于

  .then(x => all(identity)(x))

而不是

  .then(x => all(identity(x)))

在两种情况下，这些类型的推断方式会有所不同，我认为您希望在获得前者时对代码进行后一种解释。

如何实现您期望看到的错误

变体1 ：如果您重写代码，以便all()以正确的顺序获得string承诺值和identity()，那么您就可以了。我会看到错误。

Promise.resolve('kromid')
  .then(text => all(text)(identity))
  .then(([a]) => a.splita);

导致：

  

[ts] Property＆＃39; splita＆＃39;类型＆＃39;字符串＆＃39;上不存在。你是说＆＃39;分裂＆＃39;？

变体2 。重新排序重载all()函数的参数，以便它能够以您希望的方式使用它：

function all<T1, Param>(p: Param): (a1: Res<Param, T1>) => Promise<[T1]>;
function all<T1, T2, Param>(p: Param): (a1: Res<Param, T1>, a2: Res<Param, T2>) => Promise<[T1, T2]>;
function all<T1, T2, T3, Param>(p: Param): (a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>) => Promise<[T1, T2, T3]>;
function all<T1, T2, T3, T4, Param>(p: Param): (a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>) => Promise<[T1, T2, T3, T4]>;
function all<T1, T2, T3, T4, T5, Param>(p: Param): (a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>) => Promise<[T1, T2, T3, T4, T5]>;
function all<T1, T2, T3, T4, T5, T6, Param>(p: Param): (a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>, a6: Res<Param, T6>) => Promise<[T1, T2, T3, T4, T5, T6]>;
function all<T1, T2, T3, T4, T5, T6, T7, Param>(p: Param): (a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>, a6: Res<Param, T6>, a7: Res<Param, T7>) => Promise<[T1, T2, T3, T4, T5, T6, T7]>;
function all<T1, T2, T3, T4, T5, T6, T7, T8, Param>(p: Param): (a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>, a6: Res<Param, T6>, a7: Res<Param, T7>, a8: Res<Param, T8>) => Promise<[T1, T2, T3, T4, T5, T6, T7, T8]>;
function all<T1, T2, T3, T4, T5, T6, T7, T8, T9, Param>(p: Param): (a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>, a6: Res<Param, T6>, a7: Res<Param, T7>, a8: Res<Param, T8>, a9: Res<Param, T9>) => Promise<[T1, T2, T3, T4, T5, T6, T7, T8, T9]>;
function all<T1, T2, T3, T4, T5, T6, T7, T8, T9, T10, Param>(p: Param): (a1: Res<Param, T1>, a2: Res<Param, T2>, a3: Res<Param, T3>, a4: Res<Param, T4>, a5: Res<Param, T5>, a6: Res<Param, T6>, a7: Res<Param, T7>, a8: Res<Param, T8>, a9: Res<Param, T9>, a10: Res<Param, T10>) => Promise<[T1, T2, T3, T4, T5, T6, T7, T8, T9, T10]>;
function all<Param>(p: Param): (...values: Res<Param, any>[]) => Promise<any[]>;

function all<Param>(p: Param): (...values: Res<Param, any>[]) => Promise<any[]> {
  return values => Promise.all(values.map((obj: any) => obj.apply ? obj(p) : obj));
}

type Res<I, O> = ((i: I) => O | Promise<O>) | O | Promise<O>;

function identity<T>(a: T): T {
  return a;
}

Promise.resolve('kromid')
  .then(all(identity))
  .then(([a]) => a.splita);

它将实现相同的结果：

我希望，我的答案现在完整正确。 ：）

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