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Conditional types wrongly work with indexed access and object type (regression) #23800
Description
cc Anders Hejlsberg (@ahejlsberg)
TypeScript Version: 2.9.0-dev.20180501
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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:
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mhegazy commentedon May 1, 2018
I see both as
{ a: 0 }on latest and on 2.8.3 (playground)mhegazy commentedon May 1, 2018
never mind.. looks like we broke it yesterday.
ahejlsberg commentedon May 1, 2018
The immediate cause of this is #23768. However, it just exposes indexed access types to an issue that already exists with regular type parameters:
The problem is caused by some logic in
structuredTypeRelatedTothat specifically excludes checking the default{}constraint of an unconstrained type parameter against a target type ofobject. However, that logic doesn't kick in when the target is a union type that includesobject.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.