Subclassing ES6 objects with ES5 syntax.

[C. Scott Ananian]

On Wed, Apr 29, 2015 at 3:09 PM, Allen Wirfs-Brock <allen at wirfs-brock.com>
wrote:

>  class DefensivePromise extends Promise {
>   constructor(x) {
>     super(x);
>     if (new.target === DefensivePromise) {
>
           // I'm assuming this test is just to be subclass friendly and
allow subclasses to freeze later?

>       Object.freeze(this);
>     }
>   }
>   static resolve(x) {
>     switch (true) {
>        default:
>
         // I guess a do { } while(false); would work as well?

>           // assuming frozen primordial
>           if (x.constructor !== DefensivePromise) break;  //just a quick
> exit, doesn't guarantee much
>           if (!Object.isFrozen(x)) break;
>           If (Object.getOwnPropertyDescriptor(x, 'then')) break;
>           //a more subclass friendly approach would walk x's [[Prototype]]
> chain to ensure that the correct 'then' is inherited from frozen prototypes
>           if (Object.getPrototypeOf(x) !== DefensivePromise.prototype)
> break;
>           //Assert: x.then === Promise.prototype.then, now and forever
> after
>           return x;
>      }
>      // must be called on a subclass of DefensivePromise, so we don't need
> to enforce the 'then' invariant
>      If  (x.constructor ===  this) return x;   //in which case a
> constructor check is good enough
>
          // ^^ this is a mistake right?  I think this line doesn't belong.

>      return new this(r => {r(x)});
>    }
>  }
> Object.freeze(DefensivePromise);
> Object.freeze(DefensivePromise.prototype);
>

It's not clear what the `x.constructor` test is still doing in your
implementation.

But, regardless of the details of our implementations, can we agree that
"tamper proof" promises don't seem to need the [[PromiseConstructor]]
property?
  --scott
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