super.prop assignment can silently overwrite non-writable properties
Allen Wirfs-Brock
allen at wirfs-brock.com
Tue Apr 21 01:18:40 UTC 2015
On Apr 20, 2015, at 12:39 PM, Jason Orendorff wrote:
> On Mon, Apr 20, 2015 at 12:44 PM, Allen Wirfs-Brock
> <allen at wirfs-brock.com> wrote:
>>> In the spec, 9.1.9 step 4.d.i. is where `super.prop = 2` ends up, with
>>> O=X.prototype.
>>
>> 4.d.1 doesn't set the property, it just comes up with the property descriptor to use, if the `Receiver` does not already have a corresponding own property.
>>
>> 5.c+5.e checks if the corresponding own property actually exists on the `Receiver`.
>>
>> If it already exits then it does a [[DefineOwnProperty]] that only specifies the `value` attribute. This should respect the current `writable` attribute of the property and hence reject the attempt to change the value.
>
> I agree with all of this, except I don't see where the attempt is
> rejected. Since the property is configurable, I think
> [[DefineOwnProperty]] succeeds.
>
> The property is still non-writable afterwards. Only the value changes.
>
> So this isn't breaking the object invariants: the property in question
> is configurable, so it's OK (I guess) to change the value. It's just
> surprising for assignment syntax to succeed in doing it.
I think it's bogus and needs to be corrected. Not only does it allow (in weird cases for [[Set]] (ie, assignment) to change the value of a non-writable property. It also means there are cases where [[Set]] will convert an accessor property to a data property.
In combination, I think this is a serious bug that needs to be fix in the final published ES6 spec. The fix I propose is in 9.1.9 to replace Set 5.e as follows:
5.e If existingDescriptor is not undefined, then
i. If IsAccessorDescript(existingDescript), return false.
ii. If existingDescriptor.[[Writable]] is false, return false.
iii. Let valueDesc be the PropertyDescriptor{[[Value]]: V}.
iv. Return Receiver.[[DefineOwnProperty]](P, valueDesc).
Lines 5.e.i and 5.e.ii are new additions.
Thoughts?
Allen
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