Conditional types wrongly work with indexed access and object type (regression)

[falsandtru]

cc @ahejlsberg

TypeScript Version: 2.9.0-dev.20180501

Search Terms:

Code

type A<T, V> = { [P in keyof T]: T[P] extends V ? 1 : 0; };
type B<T, V> = { [P in keyof T]: T[P] extends V | object ? 1 : 0; };

type a = A<{ a: 0 | 1 }, 0>; // { a: 0; }
type b = B<{ a: 0 | 1 }, 0>; // { a: 1; }

Expected behavior:

type b = B<{ a: 0 | 1 }, 0>; // { a: 0; }

Actual behavior:

type b = B<{ a: 0 | 1 }, 0>; // { a: 1; }

Playground Link:

Related Issues:

[mhegazy]

I see both as { a: 0 } on latest and on 2.8.3 (playground)

[mhegazy]

never mind.. looks like we broke it yesterday.

[ahejlsberg]

The immediate cause of this is #23768. However, it just exposes indexed access types to an issue that already exists with regular type parameters:

function foo<T, U>(x: T) {
    let a: object = x;  // Error as expected
    let b: U | object = x;  // No error!
}

The problem is caused by some logic in structuredTypeRelatedTo that specifically excludes checking the default {} constraint of an unconstrained type parameter against a target type of object. However, that logic doesn't kick in when the target is a union type that includes object.

Instead of attempting to exclude cases, when the source is an unconstrained type variable I think we need to check if the target is an empty object type or a union containing an empty object type.