Generic parmater with union type not working as expected (not assignable to parameter of type)

[JasonMan34]

🔎 Search Terms

union, generic, assignable, not assignable to parameter of type

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about union and generics

⏯ Playground Link

https://www.typescriptlang.org/play?#code/MYewdgzgLgBFCGBrApgZSgJwJZgOYwF4YAKAB3g3gFsAuGabPASkID4YBXMAE2QDMcybgG4AUKEiwEKAHIcqAI2QZCJcpVoww8pRhYF2XXgLBCxoqAE9SyGADEuwVVZsg+cJGkw58AHzjWyG4esjrK5hLQMFTICA5gTkQAPAAqMMgAHlDIPBD2jqzEfI50KQA0MOrUdAAKFNSxyhCprADaAAwAuvrsAN6iMDAA9EMwAIIYuPI5sMEutgDkDD4w-tqKygswWHlgILDwEBBYuGDwCgA2tlAglfUx2SpzgTALpgBumwB0A8OjKS8lt48Fsdlp9jBDsdTucrnBbvNXh9Nr9igkyPcmGIAL6iUQjGAAmzbMB8ZQ5YC2ADuIAwiDyh3SGRswGy3BgVIAFjkYMB4BcLisoNyYGjWVhwPjRhMpjEwLN3Ii3mEMKDdhCoSczpdrrcqg9lDBnsSgYxcAtRA94PFgMRpF4zRUAExMURAA

💻 Code

const takeString = (param: string) => undefined;
const takeNumber = (param: number) => undefined;

type Func = typeof takeString | typeof takeNumber;

const metaFunc = <T extends Func>(func: T, param: Parameters<T>[0]) => {
  // Argument of type 'string | number' is not assignable to parameter of type 'never'.
  // Type 'string' is not assignable to type 'never'
  func(param);
}

// Type inference works as expected when calling the function
// Argument of type 'number' is not assignable to parameter of type 'string'
metaFunc(takeString, 2)

🙁 Actual behavior

The func invocation is invalid. With the error:

Argument of type 'string | number' is not assignable to parameter of type 'never'.
  Type 'string' is not assignable to type 'never'

🙂 Expected behavior

TypeScript should recognize that since the same T is used for both func and param, the invocation of func(param) is valid. Similar to how it recognizes that metaFunc(takeString, XXX) should be called with a string rather than a string | number

Additional information about the issue

Potentially related: #51587, #33912, #30581

Note that this is different than https://github.com/microsoft/TypeScript/issues/30581#issuecomment-476417974 since the same T extends Func is used for both func and param, rather than using Func for both types

[snarbies]

As defined, this is a valid invocation of metaFunc

metaFunc<Func>(takeString, 3);

// Maybe something you'd be more likely to see in the wild
const myFunc = ((arg: string) => undefined) as Func;
metaFunc(myFunc, 3);

It might not be practical in your actual use case but in your example, this workaround is available:

const metaFunc = <T extends string | number>(func: (arg: T)=>undefined, param: T) => {
    // ...
}

[JasonMan34]

I see the issue.
The workaround actually works great for me:

type Func<T extends string | number> = (arg: T) => undefined

const metaFunc = <T extends string | number>(func: Func<T>, param: T) => {
  func(param);
}

Of course in real life Func is more complicated, but as long as I can neatly wrap it in it's own type it doesn't bother me

Thanks!