Type narrowing of `super` when using `this`

Bug Report

🔎 Search Terms

super, this, class

🕗 Version & Regression Information

Nightly

⏯ Playground Link

https://www.typescriptlang.org/play?ts=5.2.0-dev.20230720#code/MYGwhgzhAECCB2BLAtmEAeAKgPmgbwChpjoBTMAFy2wAoKALRCALjiVQxwEp8iT+GAJwD2Ad2jxS4gKKCRgmgHJMjGABNhpCPArRUFCqUGKuAbj7EAvgWsFQkGABFhAc2pkAHofhqYCFGjuhPzkVACqtAxMrM5uETzB-CQQAK4ADkYAdKE0ZtAA9PnQKqTQFACeGdBMbAGc2NmU1HSqrP4c1FysAG7CiGoWSUPDhcX0pRVVEPTCKSBq0ABGpe2BEY3hka21HfE9fQP81tZAA

💻 Code

class Animal<T> {
    eat<T>(this: Animal<T>) {
        throw new Error('This doesnt matter');
    }
}

class Dog<T> extends Animal<T> {
    eat<U>(this: Dog<U>) {
        super.eat(); // The type is Animal<T>.eat<T>(this: Animal<T>): void
                     // The type should be Animal<U>.eat<U>(this: Animal<U>): void
    }
}

🙁 Actual behavior

super.eat() is typed as Animal<T>.eat<T>(this: Animal<T>): void

🙂 Expected behavior

super.eat() should be typed as Animal<U>.eat<U>(this: Animal<U>): void since this is declared as Dog<U> which is a subclass of Animal<U>.

One can workaround this by writing super.eat<U>(), but the signature then becomes goofy (it becomes Animal<T>.eat<U>(this: Animal<U>): void)

Inizia con il TypeScript Playground collegato e riproduci la differenza tra i tipi correnti e quelli previsti per super.eat() nelle classi generiche Animal e Dog. Traccia la gestione dei tipi this da parte del compilatore nell'accesso alle proprietà di super, quindi aggiungi un test di regressione e conferma che super.eat() preservi il tipo specifico di U senza richiedere un argomento di tipo esplicito.