Cannot extend mapped types #12986

Activity

1. Mohamed Hegazy (mhegazy)

   added

   SuggestionAn idea for TypeScriptAn idea for TypeScript

   In DiscussionNot yet reached consensusNot yet reached consensus

   on Dec 16, 2016
2. ahejlsberg commented on Dec 16, 2016

   

   Anders Hejlsberg (ahejlsberg)

   on Dec 16, 2016

   Member

   More actions

   This is not an easy change as it would require some deeper architectural reworking. Note that you can accomplish pretty much the same effect with the intersection type operator:

   type B = Partial<A> & { y: number };
3. rotemdan commented on Dec 16, 2016

   

   Rotem Dan (rotemdan)

   on Dec 16, 2016

   Last edited by rotemdan

   More actions

   Just wanted to add: I have considered suggesting this myself, though one thing I must mention is that even if it was made possible to inherit from computed types, that would still miss a significant use case for them as generators of "templates" for new types (rather than serving as base types). I'll try to demonstrate:

   Let's say I have an existing type of this shape:

   type SomeType = {
   	a: number;
   	b: boolean;
   	c: string;
   }

   Now I want to declare a type that has the same exact keys only each key is now mapped to a an object that has a property val with the type of the corresponding key in SomeType. Mapped types allow me to do that:

   type MyMappedType<T> = { [ P in keyof T]: { val: T[P] } };
   type Result = MyMappedType<SomeType>;

   Now I want to customize the resulting type, so the closest thing the language provides me is to "inherit" from that mapped type. Let's assume that was possible:

   interface MyInterface extends MyMappedType<SomeType> {
   	a: { val: number, x: string[], y: boolean[] };
   	b: { val: boolean, z: number };
   }

   There's only one problem here:

   I forgot to specify property c, so it "fell back" to { val: string }, however the compiler didn't tell me anything went wrong! why? because I used the wrong operator.

   What if what I want is not to "inherit" from the mapped type, but simply use it as a "model" or "constraint" for the new type?

   I have tried my best to propose a solution in #12942. It's still in an early stage though. I've submitted it at least with the hope that it might influence or inspire the typescript team to include something of that sort in the future, even if it eventually appears with a different syntax or in a more limited form.
4. kimamula commented on Dec 30, 2016

   

   Kenji Imamula (kimamula)

   on Dec 30, 2016

   Last edited by kimamula

   Contributor

   More actions

   I am trying to develop Vue.js app with TypeScript and also having a trouble because of the limitation that mapped types are not available in classes or interfaces.

   In Vue.js, you can pass props to a component (just like React) and the properties of props are injected into the properties of the component instance.
   Therefore, the declaration of the class of Vue component should be something like this, which is not possible currently.

   declare class Vue<Props> {
     [P in keyof Props]: Props[P];
   }
   
   // example usage
   class Foo extends Vue<{ message: string }> {
     render(h) {
       return <span>{this.message}</span>;
     }
   }

   Anders Hejlsberg (@ahejlsberg)'s workaround does not work in this situation.
5. Anders Hejlsberg (ahejlsberg)

   closed this as completedin #13604on Jan 21, 2017
6. Mohamed Hegazy (mhegazy)

   added

   FixedA PR has been merged for this issueA PR has been merged for this issue

   and removed

   In DiscussionNot yet reached consensusNot yet reached consensus

   on Jan 21, 2017
7. Mohamed Hegazy (mhegazy)

   added this to the TypeScript 2.2 milestone on Jan 21, 2017
8. microsoft

   locked and limited conversation to collaborators on Jun 19, 2018