TypeScript: ReturnType returns unknown if function's return type is generic argument of the function (with default value) that has not been set

🔎 Search Terms

ReturnType, generic, default

🕗 Version & Regression Information

This is a crash
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

No response

💻 Code

const func = <Return = string>(): Return {
   //
}

type ShouldBeNumber = ReturnType<typeof func<number>>; // number
type ShouldBeString = ReturnType<typeof func>; // unknown

🙁 Actual behavior

ReturnType returns unknown.

🙂 Expected behavior

ReturnType returns default type.

Additional information about the issue

This was encountered when trying to write this kind of function:


const invoker  = <Endpoint extends (...args: any) => any>(endpoint: Endpoint, ...parameters: Parameters<Endpoint>): ReturnType<Endpoint> {
    // Perform some logic before endpoint function is called

    return endpoint(...parameters);
}

About this issue

Original URL
State: closed
Created 4 months ago
Comments: 15 (2 by maintainers)

Most upvoted comments

Duplicate of #48870.

MartinJohns on Feb 21, 2024

If not removing this split competely, some expression that can be used to convert one to another might fix this issue?

Is this what you’re looking for?

const func = <Return,>(): Return => {
  throw new Error("implement me")
}
type Func<T=string,> = typeof func<T>
type ShouldBeNumber = ReturnType<Func<number>>; // number
//   ^?
type ShouldBeString = ReturnType<Func>; // string
//   ^?

https://www.typescriptlang.org/play?#code/MYewdgzgLgBAZgVzMGBeGAeASgUyggJzABoA+ACgEoAuGXfItUmAbwCgYYoALAkAdxhgcggKIE+BcgCIAlgFsADgBsc8nGFjrplNgF82UAJ6KcMAGJJgGACqpoBWWADmZNFxM4QceFdulDTxgAZW4QBGUAEwAhHAA5BHkAIxwCd3pCMBtPDEtkDDBElIJSUgBuGAB6SqEi1LZqzhgAPQB+QNMQsIiYnGCoRxd0vEzs01yrcqqahydnBprONqA

rotu on Feb 21, 2024

That exhibits the same distinction as

type Fun1<T = string> = () => T;
type Fun2 = <T = string>() => T;

One is a generic type (i.e. a type-function) that resolves to a concrete, non-generic function type. The other is the type of a generic function, whose return type is not known until it’s called.

Note that this is the same reason you can’t write Fun2<number>.

fatcerberus on Feb 21, 2024

I do not think this is working as indented in sense that it might be an intentional side effect but it is frankly stupid to claim that this is “proper” behaviour that should be upheld.

If this is working

type ExtractDefault<T> = T extends infer U ? U : never;
type MyType<T = number> = T;
type DefaultType = ExtractDefault<MyType>; // number, NOT unknown

I see no reason why it should not work with functions.

cts-tradeit on Feb 21, 2024