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Function parameter type inference error #59949

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lengyuxuan opened this issue Sep 12, 2024 · 4 comments
Closed

Function parameter type inference error #59949

lengyuxuan opened this issue Sep 12, 2024 · 4 comments

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@lengyuxuan
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🔎 Search Terms

Function parameter

🕗 Version & Regression Information

  • This changed between versions ______ and _______
  • This changed in commit or PR _______
  • This is the behavior in every version I tried, and I reviewed the FAQ for entries about _________
  • I was unable to test this on prior versions because _______

⏯ Playground Link

https://www.typescriptlang.org/zh/play/?#code/KYOwrgtgBAKgngB2FA3gWAFBSgQQDSbYBCBGAvppgJYgAuwATgGYCGAxsvEgLIsKqEoAbS7AAdDgC6ALigBnWgxoBzANyCRicURlRwEAEaN15ShlpaoADQA8MKMAAe9EABM5sLQD4oAXmyivAgikiaYTGAgbLRUAPYgUBFRdg7OoO6eSF4AFBZIsjB4qS6usrYwXgCUAlhQVExQuZa+LZniONXotdgA9D1QAOoAFnBQAEROJWNQechUHiCxtPKKKlAA-IK9-dkILAwsEMD0DNWT6bIKSiDKUAA+epBGDPdQBrGxADbALCBbUGx4nIvuJPrFlE0kEVzm5KiZsBRTBhwpE2NlRBIigByLFwlFRdFaMQkKAARjhQA

💻 Code

enum Type {
  A,
  B,
}

interface TypeMap {
  [Type.A]: string;
  [Type.B]: number;
}

type X<T extends Type> =  TypeMap[T];

function func<T extends Type>(type: T, extend: X<T>) {
  if (type === Type.A) {
    // Why "extend" type is not string ?
    // (parameter) extend: string | number | boolean
    console.log(type, extend);
  }
}

func(Type.A, '');
func(Type.B, 1);

🙁 Actual behavior

In the example, Why "extend" type is not string ?

🙂 Expected behavior

The "type" constrained to "Type.A",so "extend" shuld be "string".

Additional information about the issue

No response

@MartinJohns
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Contributor

Because checking what value type is does not have any relevance for what type the value of extend is.

This is a perfectly valid call: func<Type.A | Type.B>(Type.A, 1);

@lengyuxuan
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Author

Because checking what value type is does not have any relevance for what type the value of extend is.

This is a perfectly valid call: func<Type.A | Type.B>(Type.A, 1);

How should I constrain the type of "extend" ? Do you have any suggestions ? Thank you so much !

@jcalz
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jcalz commented Sep 12, 2024

Duplicate #33014

For now, if you need this behavior you either need to rewrite in terms of #47109 and use generic indexing instead of control flow anlaysis:

function func<T extends Type>(type: T, extend: X<T>) {
    const x: { [K in Type]: (type: K, x: X<K>) => void } = {
        [Type.A](type, extend) {
            console.log(type, extend);
            //                ^? (parameter) extend: string
        },
        [Type.B](type, extend) {

        }
    }
    x[type](type, extend);
}
func(Type.A, '');
func(Type.B, 1);
func(Math.random() < 0.99 ? Type.A : Type.B, 1); // still allowed, #47109 technically unsound this way

or, you need to use control flow analysis on a discriminated union and not use generics:

function func(...[type, extend]: { [T in Type]: [type: T, extend: X<T>] }[Type]) {
    if (type === Type.A) {
        console.log(type, extend);
        //                ^? (parameter) extend: string
    }
}

func(Type.A, '');
func(Type.B, 1);
func(Math.random() < 0.99 ? Type.A : Type.B, 1); // error

Playground link to code

@lengyuxuan
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Author

Duplicate #33014

For now, if you need this behavior you either need to rewrite in terms of #47109 and use generic indexing instead of control flow anlaysis:

function func<T extends Type>(type: T, extend: X<T>) {
    const x: { [K in Type]: (type: K, x: X<K>) => void } = {
        [Type.A](type, extend) {
            console.log(type, extend);
            //                ^? (parameter) extend: string
        },
        [Type.B](type, extend) {

        }
    }
    x[type](type, extend);
}
func(Type.A, '');
func(Type.B, 1);
func(Math.random() < 0.99 ? Type.A : Type.B, 1); // still allowed, #47109 technically unsound this way

or, you need to use control flow analysis on a discriminated union and not use generics:

function func(...[type, extend]: { [T in Type]: [type: T, extend: X<T>] }[Type]) {
    if (type === Type.A) {
        console.log(type, extend);
        //                ^? (parameter) extend: string
    }
}

func(Type.A, '');
func(Type.B, 1);
func(Math.random() < 0.99 ? Type.A : Type.B, 1); // error

Playground link to code

Good job

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3 participants