I filed one on us: https://code.google.com/p/v8/issues/detail?id=4087

On Thu, May 7, 2015 at 4:25 PM Allen Wirfs-Brock <allen at wirfs-brock.com>
wrote:

>
> On May 7, 2015, at 12:50 PM, Francisco Tolmasky wrote:
>
> > In the existing implementations I’ve tried, it appears I can’t do this:
> >
> > class SuperFunction extends Function { }
> >
> > (also tried with constructor(str) { super(str) })
> >
> > It “works”, but the resulting new SuperFunction(“return 5”) is just a
> Function, not a SuperFunction. Is Function meants to be an exception to the
> subclassing built-ins, or should it also work?
>
> Nope, it's supposed to work. `Object.getPrototypeOf(new SuperFunction("")`
> should be SuperFunction.prototype and `new SuperFunction("") instanceof
> SuperFunction` should be true.
>
> You need to file a bug report on the implementations where you tried it.
> Sounds like they still have some work to do
>
> Allen
>
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I filed one on us: https://code.google.com/p/v8/issues/detail?id=4087