Suggestion: allow `infer T` in type parameter constraint of `type` statement

Search Terms

suggestion infer extends type parameter constraint

Suggestion

Allow infer T in the right of extends of a type statement.

Examples

This suggestion allows following code:

type ReturnType<T extends (...args: any) => infer R> = R;

Which behaves similarly as:

// this code is taken from lib.es5.d.ts
type ReturnType<T extends (...args: any) => any> = T extends (...args: any) => infer R ? R : any;

Use Cases

As demonstrated in the example, we sometimes use conditional types with the sole purpose of use of infer T.
We have to repeat some piece of code if we want to constraint what types can be passed to ReturnType.
So this suggestion provides an abbreviated syntax for this use case.

Open questions:

• Should union distribution occur for the new syntax?
  • My opinion is yes to align with existing conditional types
  • hmm, but then just replacing any with infer _ would change the behavior...
• Should function type parameters also allow infer T?
  • No strong opinion

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.

Aucun fichier source, test ou point d’entrée n’est nommé. Commencez par examiner la syntaxe infer T proposée et sa comparaison de types conditionnels, puis résolvez les questions en suspens concernant la distribution des unions et les paramètres de type des fonctions. C’est terminé lorsque la syntaxe et sa sémantique sont suffisamment spécifiées pour être implémentées et testées.