It's possible to create a callable `never`

[mkantor]

Bug Report

🔎 Search Terms

call never, callable intersection, intersection call signature, intersection discriminant

🕗 Version & Regression Information

• This changed between versions 3.8.3 and 3.9.7 (and still happens in 4.1.5 and 4.2.0-beta).

⏯ Playground Link

Playground link with relevant code

💻 Code

declare const f: { (x: string): number, a: "" } & { a: number }
const result: "bad" = f()

🙁 Actual behavior

The above code typechecks! f's type is displayed as never, but it behaves like (...args: any[]) => any.

🙂 Expected behavior

f should either be a real never (and therefore not be callable), or be callable with the specified call signature ((x: string) => number). Probably the former?

---

This occurs when intersecting a type that has a call signature with another incompatible type, and only if the incompatibility involves a literal type (i.e. when it's a discriminant).

Based on the behavior and affected versions, this is likely related to #36696.

Elaborated playground with commentary and more examples

[Maximilian-Seitz]

This bug also occurs without a call signature in the original portions of the 'never' type.

So this compiles (Playground Link):

declare const f: { a: '' } & { a: 1 }
const result: "bad" = f()

While this doesn't:

declare const f: never
const result: "bad" = f()