type alternatives in arguments become function alternatives and are then inferred to a never argument.

[sparecycles]

🔎 Search Terms

never alternative types ts2345

🕗 Version & Regression Information

• This is the behavior in every version I tried.

⏯ Playground Link

https://www.typescriptlang.org/play/?ts=5.7.3#code/GYVwdgxgLglg9mABANzjAJgeQE4Dk5QA8AKogLwproB8AFAJQBciA3osM6QKYAeUXYdAGdKGRAH5EDctVHpEzWgENsAc070ZcxAF9WAKESJsXKCGxIwIADbWAhAG59O-fqgBPAA5dEAQQBC5IgARL7BiAA+If7BTvoQCEJQiADuMFAAFgDKMAC2ntZcxF4+FKgYOPhEocF09HFpmTn5hcXeAHTAtDX1rglgSanpGQBqVG2lcpUEhOU0DE6No+MlnQt9iclLvtb8FkqwyEUlQXPTRAF1DcM7e2AHMEcTa73bu1z7h8cdXT1xAPT-RAebxwYBDTK3D73L4TZhsDhSZRqZg9LRzTRRWjI9TRYKaMiyDG6fSA4ElMEQjJQz6Pb5cTqKFS4sBcI7YAlEqhOIA

💻 Code

function voidOrNot<T = void>(): { f: T extends void ? () => void : (arg: T) => void } {
  return null!;
}

const withSimpleType = voidOrNot<"A">();
withSimpleType.f("A"); // OK

const withVoidType = voidOrNot<void>();
withVoidType.f(); // OK

type AB = "A" | "B";

const withAlternativeType = voidOrNot<AB>();

withAlternativeType.f(); // error
withAlternativeType.f("A"); // error

🙁 Actual behavior

---

---

🙂 Expected behavior

f(arg: AB) => void is not expanded to (f(arg: A) => void) | (f(arg: B) => void)... or that it's still callable if it is.

Additional information about the issue

please feel free to suggest a better issue title, I'm having trouble picking a good name today!

[Andarist]

You might have wanted to use a non-distributive conditional type there:

function voidOrNot<T = void>(): { f: [T] extends [void] ? () => void : (arg: T) => void } {
  return null!;
}

[sparecycles]

@Andarist that solved my issue, thanks!

---

Thinking about this.... I left out that the oldest version of typescript (3.3.3) in the Playground changes "A"|"B" to "A"&"B"

(which was also wrong).

---

If conditional types are meant to distribute, wouldn't transforming (arg: "A"|"B") => void to ((arg: "A") => void) & ((arg: "B") => void) be the correct way to do it (with & instead of |)?

[sparecycles]

Searched more, looks like this is a duplicate of, or related to

• Conditional type is mangling type signature when distributing a union across a function signature #30378 (fixed by the above suggestion?)
• Type inference for conditional expands [A | B] to [A] | [B] #29535 (not fixed by the above suggestion)

[Andarist]

If conditional types are meant to distribute, wouldn't transforming (arg: "A"|"B") => void to ((arg: "A") => void) & ((arg: "B") => void) be the correct way to do it (with & instead of |)?

That might be what you want to get in certain situations but the compiler can't predict your intention. A union distributes into another union

[sparecycles]

I'm not convinced this is the best language decision, but thanks again for the tight workaround!