Allow inference of class generics when declaring an interface

Search Terms

interface, inference, extending class

Suggestion

I'd like to be able to strongly type an interface such that when it is used on a class that has a generic I can infer the type of that generic to then use in the declaration of the interface.

Use Cases

I want to use this within a library that exposes both an abstract class and an optional interface that developers can opt-in to use. The abstract class has one generic parameter, and in order to correctly type the interface without duplicate generic declaration I want to be able to infer the abstract class' generic type.

Currently the inheriting class declarations must duplicate their generic type declaration.

This issue is similar to https://github.com/microsoft/TypeScript/issues/26242 I think, but I don't see how the resolution of that issue will fix this one.

Examples

abstract class Base<T> {
    // implementation omitted for brevity
}

// ⬇️this is the proposed new syntax ⬇️
interface Remap<T> extends Base<infer U> {
    remap(input: U): T;
}

class Foo extends Base<string> implements Remap<number> {

    public remap(input: string): number {
        return input;
    }

}

Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

Issue では、ソースファイル、テスト、またはコンパイラのエントリポイントが特定されていません。まず提案されている TypeScript 構文と関連する issue #26242 を読み、次にジェネリック継承とインターフェイス宣言が現在どのようにチェックされているかを追跡してください。チームが意図した推論動作を定義して実装し、示されている例とそのエッジケースをカバーできれば完了です。