Optional without undefined union

Sorry if this is an ignorant question but is it possible to create an object with an optional key without the | undefined?

i.e. this

type Foo = {
  bar?: string;
}

instead of the

type Foo = {
  bar?: string | undefined;
}

You get when using .optional().

[colinhacks]

Nope that’s not possible. They’re treated identically by TypeScript so there’s no reason for it. You can see that the | undefined gets automatically added by TS during any operation on the type, even trivial ones:

type Foo = {
  bar?: string;
};

type id<T> = T;
type NewFoo = id<Foo>; // { bar?: string | undefined }

[scotttrinh]

Yeah, exactOptionalPropertyTypes is a pretty new feature, since it came out in 4.4. Might take some time to implement and require some care. I feel like maybe we discussed having it as a separate method to avoid breaking backwards compatibility, but I can’t seem to find the issue now.