type A = "a" | "b";
type B = "c" | "d";
type C<Type extends A> = Type;
type D<Type extends B> = Type;
type Auto<Type extends (A|B)> = Type extends A ? C<Type> : D<Type>; //It throws error!
//Type 'Type' does not satisfy the constraint 'B'.
Auto type has generic. Type is A|B, same as "a" | "b" | "c" | "d". And A is equal to "a" | "b". But why I can't use Type extends A ? C<Type> : D<Type>? D<Type> throws error "Type 'Type' does not satisfy the constraint 'B'.".
This is an open issue; see microsoft/TypeScript#23132. You may want to give that issue a 👍 or describe your use case if you think it's compelling. Not sure if it will ever be changed. For now, though, conditional types more or less ignore any generic constraint and the workaround is to use additional and possibly redundant checks:
type Auto<T extends (A | B)> = T extends A ? C<T> : T extends B ? D<T> : never
That should behave the way you want it to. Okay, hope that helps; good luck!
