Description

fangjin (lengyuxuan)

opened on Sep 12, 2024

Issue body actions

🔎 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

Activity

1. MartinJohns commented on Sep 12, 2024

   

   Martin Johns (MartinJohns)

   on Sep 12, 2024

   Contributor

   More actions

   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);
2. lengyuxuan commented on Sep 12, 2024

   

   fangjin (lengyuxuan)

   on Sep 12, 2024

   Author

   More actions

   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 !
3. jcalz commented on Sep 12, 2024

   

   Joe Calzaretta (jcalz)

   on Sep 12, 2024

   Contributor

   More actions

   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

   Reacted by Ryan Cavanaugh and fangjin
4. lengyuxuan commented on Sep 13, 2024

   

   fangjin (lengyuxuan)

   on Sep 13, 2024

   Author

   More actions

   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
5. fangjin (lengyuxuan)

   closed this as completedon Sep 13, 2024
6. fangjin (lengyuxuan)

   reopened this on Sep 13, 2024
7. fangjin (lengyuxuan)

   closed this as completedon Sep 13, 2024
8. microsoft

   locked as resolved and limited conversation to collaborators on Oct 22, 2025

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