Accessing base class field through super should be disallowed

[Josh-Cena]

Bug Report

🔎 Search Terms

Super, class, field, property

Related: #45712, #42215, which are about allowing legitimate accesses, while this is about disallowing illegal access.

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about classes

⏯ Playground Link

Playground link with relevant code

💻 Code

class A {
  a = 1;
}

class B extends A {
  calc() {
    // super.a has type number, but should be disallowed
    const a = super.a + 1;
  }
}

new B().calc();

🙁 Actual behavior

I'm allowed to access super.a, which is undefined at runtime.

🙂 Expected behavior

It should be forbidden, because super is an alias for instance.__proto__.__proto__, which is A.prototype, not an instance of A.

[Josh-Cena]

I think this is due to the fact that TS has no concept (afaik) of the prototype chain. It thinks super means any instance of A.

[MartinJohns]

Duplicate of #35314. Used search terms: super in:title

[Josh-Cena]

Oh, oops, used too many keywords in my query. Thanks for pointing out.