Making "super" work outside a literal?

Brendan Eich brendan at
Sun Jun 26 12:48:45 PDT 2011

On Jun 26, 2011, at 10:54 AM, Axel Rauschmayer wrote:

> This would only work inside an object literal

or in a method in a class.

> (similar to |super|, where you need to know about |here|, the owning object).
> Additionally, Allen’s Object.defineMethod already has a parameter with the method name, so it would work there, too.

Allowing 'super' in any function, requiring Object.defineMethod to be used with an object and a name for that function to be callable and survive its first super reference, is plausible and as you note, there's a name to burn into the function if necessary.

But I still wonder if we wouldn't be better off restricting where super can occur. I can't prove it, but we are following in the universal-'this' footsteps (but with static or else Object.defineMethod binding). That sounds a warning bell in my head.


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