Microsoft/TypeScript

TypeScript can't infer type of default parameters

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#59,643 创建于 2024年8月15日

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 (2 评论) (0 反应) (0 负责人)TypeScript (6,726 fork)batch import
Domain: check: Type InferenceHelp WantedPossible Improvement

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描述

🔎 Search Terms

infer default parameters function wrapper function factory

🕗 Version & Regression Information

  • This is the behavior in every version I tried, and I reviewed the FAQ
    • I tried down to version 4.0.3 (since in the earlier version Generator type is not generic)

⏯ Playground Link

https://www.typescriptlang.org/play/?#code/CYUwxgNghgTiAEAzArgOzAFwJYHtVNQBEQYsA3EAHgEF4QAPDEVYAZ3jQGtUcB3VANoBdADTwASgD4AFACh4CggC540gHQbYAc1YrqASngBeSfADizElAw4YlLj35ipI2fpUO++APQAqX-AAtjhgnCDA8HAYyDD4vt4A3LKyiEQk5CDSKOjYePC+0lDG8AAMYgBGxQDkVYYA3vLwYHisOBAgahA4WtINivBQrvDe3vAAegD8jQrl8iPjUwoAvvpJK0lAA

💻 Code

declare function fnDerive<A extends unknown[], R>(
    fn: (...args: A) => Generator<unknown, R>,
): unknown /** mocked return */;

fnDerive(function *(a = 0, b = '') {
  console.log({
    a,
 // ^?
    b
 // ^?
  });
});

🙁 Actual behavior

The type of a and b is unknown.

🙂 Expected behavior

The type of a and b should be number and string

Additional information about the issue

I was working with a library gensync when I hit this issue, and I extracted the minimum reproduction out of it.

Note that if I don't use default parameters, typescript can infer the a and b types correctly:

declare function fnDerive<A extends unknown[], R>(
    fn: (...args: A) => Generator<unknown, R>,
): unknown /** mocked return */;

fnDerive(function *(a?: number | undefined, b?: string | undefined) {
  console.log({
    a,
 // ^?
    b
 // ^?
  });

  a ??= 0;
  b ??= '';
  console.log({
    a,
 // ^?
    b
 // ^?
  });
});

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