TypeScript: exclude only exact {} from union

Question

I have a union type like:

type T = {} | ({ some: number } & { any: string })

Then how can I narrow this type to the latter? Of course this didn't work:

type WithEntries = Exclude<T, {}>

resulting in never.

Is this possible?

`{}` is any object. What you are trying to do has no sense. What are you trying to achieve really? — Nurbol Alpysbayev, Nov 30 '18 at 22:44
T is yielded by type inference. I have an object property whose type is infered as something like T above, and accessing `some` emits an error like "Property 'some' does not exist on type '{}'." — yuhr, Nov 30 '18 at 22:50

score 3 · Accepted Answer · answered Nov 30 '18 at 23:10

Here you go

type T = {} | ({ some: number } & { any: string })

type X<T> = T extends {} ? ({} extends T ? never : T) : never;

type WithEntries = X<T>; //  { some: number; } & { any: string; }

The first condition 'distributes' parts of the union type, so that the second condition can 'filter out' empty type by converting it to never, the result is union of never | (non-empty parts of T), and union never | P is just P for any P.

The idea comes from this answer.

Ooh looks pretty complicated but suits my use case. Thanks! – yuhr Nov 30 '18 at 23:21 — yuhr, Nov 30 '18 at 23:21

TypeScript: exclude only exact {} from union

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