super in private method

[morten-krogh]

The following example contains a call to a super class method in a base class using the keyword super.

class B {
	getValue() {
		return 5;
	}
}

class D extends B {

	pub() {
		return this.#priv();
	}

	#priv() {
		return super.getValue();
	}
}

const d = new D();
const value = d.pub();
console.log('value =', value);

Building this to targets, like es2016, without private methods lead to a free standing function with the super keyword.

var __accessCheck = (obj, member, msg) => {
  if (!member.has(obj))
    throw TypeError("Cannot " + msg);
};
var __privateAdd = (obj, member, value2) => {
  if (member.has(obj))
    throw TypeError("Cannot add the same private member more than once");
  member instanceof WeakSet ? member.add(obj) : member.set(obj, value2);
};
var __privateMethod = (obj, member, method) => {
  __accessCheck(obj, member, "access private method");
  return method;
};
var _priv, priv_fn;
class B {
  getValue() {
    return 5;
  }
}
class D extends B {
  constructor() {
    super(...arguments);
    __privateAdd(this, _priv);
  }
  pub() {
    return __privateMethod(this, _priv, priv_fn).call(this);
  }
}
_priv = new WeakSet();
priv_fn = function() {
  return super.getValue();
};

This output is not valid javascript.

[hyrious]

Note that tsc does the same:
Playground Link

class A extends Error {
  #f() { return super.g(); }
}

becomes

var _A_instances, _A_f;
class A extends Error {
    constructor() {
        super(...arguments);
        _A_instances.add(this);
    }
}
_A_instances = new WeakSet(), _A_f = function _A_f() { return super.g(); };

[evanw]

I believe the relevant TypeScript issue is this one: microsoft/TypeScript#44515