3.6.4 Partial field matching does not work for conditional type

[crisslion]

TypeScript Version:

3.6.4

Search Terms:
Partial field conditional matching
3.6.4 conditional partial
Code

export type QueryType = (
  { __typename?: 'TypeOne' } & { a: boolean }
) | (
  { __typename?: 'TypeTwo' } & { a: boolean }
) | (
  { __typename?: 'TypeThree' } & { a: boolean }
);

type PickFromUnion<Union, Typename> =
  Union extends { __typename?: Typename } ? Union : never;

type FirstType = PickFromUnion<QueryType, 'TypeOne'>;

Expected behavior:
FirstType would be of type { __typename?: 'TypeOne' } & { a: boolean }
Actual behavior:
FirstType is of type QueryType

Playground Link:
3.6.3 where it works
http://www.typescriptlang.org/play/?ts=3.6.3#code/KYDwDg9gTgLgBDAnmYcCKBXYVEBVmoC8cAFAFBxwDecA+rUigHYCGAtsAPwBccA5PhQB5JsD5wAvnABk1OC14AjCBAA2wFk0lkAlHAA+pCnPqNgrDj36DguAO4RxU2TQVxlajVom6DRyjSmBBZcvAIEuAAWUMBikjJybh7qmto6ANxkZGZwAAoAlgDGANYAYlAQbACqTPkQTAA8NXVMADRwNiEAfHCExs31cKAw5gAmAM4mDMHsoR0zHPGccANavKIAbtiZ2QRwpflQ4zA2vXlFZRXVtfUNmNh4BO3hwqJ8XelAA

3.7 where it does not work (3.6.4 not provided in playground)
http://www.typescriptlang.org/play/?ts=3.7-Beta#code/KYDwDg9gTgLgBDAnmYcCKBXYVEBVmoC8cAFAFBxwDecA+rUigHYCGAtsAPwBccA5PhQB5JsD5wAvnABk1OC14AjCBAA2wFk0lkAlHAA+pCnPqNgrDj36DguAO4RxU2TQVxlajVom6DRyjSmBBZcvAIEuAAWUMBikjJybh7qmto6ANxkZGZwAAoAlgDGANYAYlAQbACqTPkQTAA8NXVMADRwNiEAfHCExs31cKAw5gAmAM4mDMHsoR0zHPGccANavKIAbtiZ2QRwpflQ4zA2vXlFZRXVtfUNmNh4BO3hwqJ8XelAA

[ahejlsberg]

This was broken by #33213. Simple repro:

declare let x1: { __typename?: 'TypeTwo' } & { a: boolean };
let y1: { __typename?: 'TypeOne' } & { a: boolean} = x1;  // No error!

Before #33213 this was an error, after it isn't.

[sandersn]

This change happened in 3.6.4, so I'm going to move this into 3.8.

[sandersn]

@crisslion In the meantime, you can work around this by making the discriminant required.

Actually, why was the discriminant optional in the first place?

[sandersn]

For the minimal repro:

1. Is { t?: "Two" } & { a } assignable to each type in { t?: "One" } & { a }?
2. Some type in { t?: "Two" } & { a } must be assignable to { t?: "One" } and some type in { t?: "Two" } & { a } must be assignable to { a }.
3. { a } is assignable to { t?: "One" } (because we are checking an intersection, weak type checks are turned off) and { a } is assignable to { a }.

So { t?: "Two" } & { a } is assignable to each type in { t?: "One" } & { a }.

[crisslion]

@sandersn We are generating typescript types from graphQL types so we are using __typename to pick specific type from union. __typename is optional because then don't have to make it optional/omit it when we are using the type.

I guess more precise data type we want to pick from is:

export type QueryType = (
  { __typename?: 'TypeOne' } & { baz: boolean }
) | (
  { __typename?: 'TypeTwo' } & { foo: boolean }
) | (
  { __typename?: 'TypeThree' } & { bar: boolean }
);

So only convenient way to select the right type is using type name.
At the moment as a workaround we reverted to previous version we had 3.6.2

[sandersn]

__typename is optional because then don't have to make it optional/omit it when we are using the type.

I don't understand this part. Can you show me an example of how you omit it when using the type? How would it look if the property weren't optional?

[sandersn]

Fix is available at #34789