Mixin class with type param seems to lose information about default type parameter, causing "base constructors must all have the same return type"

🔎 Search Terms

Mixin class with type param seems to lose information about default type parameter, causing "base constructors must all have the same return type"

Related:
https://github.com/microsoft/TypeScript/issues/37142
https://github.com/microsoft/TypeScript/issues/40110

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about mixins and type parameters

⏯ Playground Link

https://www.typescriptlang.org/play?ts=5.2.2#code/CYUwxgNghgTiAEAXAngBwQYQPYDsDOiMArmIljACpogA8FAfPALzxQBGBMUp8OIA7vAAUASmaMKAbgBQ0sLgLwAQrHYQQAMWbw6KvAhAAPRCBzA88bPkIkylajQDybAFbhE9ekL0gAXPAofMSZGAG9peEio9k5uRHhIKDwLFS42dRpU7U4ASxwAc0YjEzMUpIRwqKqotlghQ39cgrFQ+DhEIhgceENJeABfCOrB6raQDq7lVXSQGX6ZOWhkqeB4YtNzKbT1DSFnN1IWwcWkiwBxLCxV9dKttU0913cRGibC+FD+oA

💻 Code

declare type ConstructorType<T> = abstract new () => T;

const BarableF = <TBase extends ConstructorType<Object>>(Base: TBase) => {
      abstract class Barable<Bar = string> extends Base {
          bar(x: string) { return x; }
      }
      return Barable;
};

class Bad extends BarableF(Object) {}

class Good extends BarableF(Object)<string> {}

🙁 Actual behavior

The line class Bad extends BarableF(Object) {} fails because "Base constructors must all have the same return type". If I hover over BarableF, I see the type

const BarableF: <ObjectConstructor>(Base: ObjectConstructor) => ((abstract new <**Bar**>() => BarableF<ObjectConstructor>.Barable<Bar>) & {
    ...;
}) & ObjectConstructor

where notably Bar has lost its default type param, which might be causing the issue.

🙂 Expected behavior

I would expect Bad to type-check without a problem, because in the Good example below it, you see that providing a type argument that is the same as the default for the type param causes things to work.

Additional information about the issue

No response

Commencez par le TypeScript Playground lié et réduisez les exemples Bad et Good à un cas de régression. Étudiez l'inférence du constructeur des mixins génériques et la raison pour laquelle le paramètre de type par défaut de Bar est perdu ; le travail est terminé lorsque Bad est validé par le système de types sans argument string explicite, tandis que Good continue de fonctionner. Ajoutez le cas à la zone de tests du compilateur concernée une fois le comportement localisé.