base value of super reference indicates the prototype of the current derived class

# Yusuke Suzuki (13 years ago)

Hello everyone,

I think that I found a issue on super reference of current draft. In the current draft, Super Reference is made with base value is result of calling GetHomeObject of env. But, HomeObject is prototype of derived class, we cannot find method of superclass.

So for example,

class Base { say() { console.log('base'); } };

class Derived extends Base { say() { console.log('derived'); super(); } };

let instance = new Derived;

instance.say();

then, because base of super reference becomes Derived.prototype, console.log('derived') is executed recursively and max recursion limit exceeded error is thrown.

So I suggest that using baseValue.[[Prototype]] for base of super reference.

If I missed something, I would appreciate it if you would inform me.

, Yusuke Suzuki

Hello everyone,

I think that I found a issue on super reference of current draft.
In the current draft, Super Reference is made with base value is result of
calling GetHomeObject of env.
But, HomeObject is prototype of derived class, we cannot find method of
superclass.

So for example,

class Base {
  say() {
    console.log('base');
  }
};

class Derived extends Base {
  say() {
    console.log('derived');
    super();
  }
};

let instance = new Derived;

instance.say();


then, because base of super reference becomes Derived.prototype,
console.log('derived') is executed recursively and max recursion limit
exceeded error is thrown.

So I suggest that using baseValue.[[Prototype]] for base of super reference.

If I missed something, I would appreciate it if you would inform me.

Regards,
Yusuke Suzuki
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