If I understand the spec correctly, destructuring works as follows:
```js
let {} = undefined; // OK???
let {x} = undefined; // TypeError
let [] = undefined; // TypeError
let [y] = undefined; // TypeError
```
Destructuring `undefined` (or `null`) via `{}` does not throw an exception (as
per first rule inside [1]). Shouldn’t it throw one?
Thanks!
Axel
[1]
https://people.mozilla.org/~jorendorff/es6-draft.html#sec-runtime-semantics-destructuringassignmentevaluation
--
Dr. Axel Rauschmayer
[email protected]
rauschma.de
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