A behavior of conditional types is changed (regression) #30047

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1. falsandtru commented on Feb 22, 2019

   

   falsandtru (falsandtru)

   on Feb 22, 2019

   ContributorAuthor

   More actions

   #30010 may fix this.
2. Ryan Cavanaugh (RyanCavanaugh)

   assigned

   Wesley Wigham (weswigham)

   on Feb 27, 2019
3. Ryan Cavanaugh (RyanCavanaugh)

   added

   Needs InvestigationThis issue needs a team member to investigate its status.This issue needs a team member to investigate its status.

   on Feb 27, 2019
4. jack-williams commented on Feb 27, 2019

   

   Jack Williams (jack-williams)

   on Feb 27, 2019

   Last edited by jack-williams

   Collaborator

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   Another smaller repro:

   interface Foo<T> {
       prop: T extends unknown ? true : false;
   }
   
   const foo = { prop: true } as const;
   const x: Foo<1> = foo;
   const y: Foo<number> = foo;
   const z: Foo<number> = x; // error

   A smaller repro.

   type Children =
       | Children.Void
       | Children.Text
       | Children.Collection
       | Children.Record;
   
   namespace Children {
       export type Void = undefined;
       export type Text = string;
       export interface Collection extends ReadonlyArray<El> { }
       export type Record = { [field: string]: El; };
   }
   
   interface El<
       C extends Children = Children,
       > {
           children: Relax<C>;
       }
   
   type Relax<C extends Children> = C extends Children.Text ? Children.Text : C;
   
   const a: El = 1 as any as El<Children.Void>; // error

   I think the issue here is that variance markers are marking C in Relax as invariant. This is probably caused by #27697 and the extension to variance computation.

   Even though TypeScript never ends up relating two conditional types in a way the requires invariance of the check types, the effect of variance computation makes the assignment illegal.