TypeScript: Using conditional type without type variables

Question

Given a union of:

type Status =
    | "Loading"
    | "Success"
    | "Error"

I was expecting this to work:

type Success = Status extends "Success" ? Status : never;
const s: Success = "Success";

However, the above results in

Type 'string' is not assignable to type 'never'.

Whereas, if I use what I thought was an equivalent:

type E<T, U> = T extends U ? T : never;
type Success = E<Status, "Success">;

const s: Success = "Success";

It compiles without a problem. Now, I know that the E type I've created is pretty much how Extract works. I'd just like to understand, what makes it behave differently in the above two cases.

Oblosys · Accepted Answer · 2021-02-09 13:30:01Z

2

It's because the conditional type distributes over the union, so E<"Loading" | "Success" | "Error", "Success"> is resolved as

E<"Loading", "Success"> | E<"Success", "Success"> | E<"Error", "Success">
// "Success"

and not as

"Loading" | "Success" | "Error" extends "Success" ? "Loading" | "Success" | "Error" : never
// never

For more information, see the TypeScript handbook section on Distributive conditional types.

answered Feb 9, 2021 at 13:30

Oblosys

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