Subclassing ES6 objects with ES5 syntax.

[Allen Wirfs-Brock]

On Apr 29, 2015, at 12:24 PM, C. Scott Ananian wrote:

> 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.

Do you mean the first or second occurrence?

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

I agree with you, but I'm not the Promise champion.

Regardless, too late for ES2015.  It will have to proposed as an ES2016 change.

Allen
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