support match for type infer

Search Terms

Suggestion

As a suggestion, there is any plans to support match keyword in type inference to simplify the type destruction and T extends P test and make code more expressive.
Support match in type inference.

Use Cases

type Infer<T> = match T {
  { prop: infer P } => P
  [infer P] => P
  _ => never // default
}

The above is equal to:

type Infer<T> = T extends { prop: infer P }
  ? P
  : T extends [infer P]
    ? P
    : never

## Checklist

My suggestion meets these guidelines:

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

The issue names no files, tests, or entry points. Start by comparing the proposed match-based type inference with the conditional-type examples in the description; done means the syntax, inference behavior, and default case have an agreed design and corresponding implementation plan.