Tab Atkins Jr. (2013-05-13T18:15:38.000Z)
github at esdiscuss.org (2013-07-12T02:27:21.454Z)
On Mon, May 13, 2013 at 11:08 AM, Andreas Rossberg <rossberg at google.com> wrote: > The case I was talking about is simply this: > > ```js > function* g() { > yield* [1, 2] > } > > var o = g() > o.send(undefined) > o.send(5) // what does this mean? > ``` > > But I suppose the answer is that the sent value is just dropped on the > floor, as per the iterator expression interpretation you gave in the > other post. Makes sense, I guess. Yes, because that code is equivalent to: ```js function* g() { yield 1; yield 2; } ``` Using `.send()` instead of `.next()` makes no difference to this code. Using `.throw()` should just percolate the error up to the `yield*` expression, as Allen suggested.