Right, it doesn’t look like one needs to know the returned value when
forwarding `return()`.
But: you need to guard against other ways of reaching `finally`. Maybe like
this:
```js
function* take(n, iterable) {
let iterator = iterable[Symbol.iterator]();
n = +n; // make sure it's a number, so that n>0 does never throw
let forwardReturn = true;
try {
while (n > 0) {
let item = iterator.next();
if (item.done) {
forwardReturn = false;
return item.value;
}
yield item.value;
n--;
}
forwardReturn = false;
} catch (e) {
forwardReturn = false;
iterator.throw(e);
} finally {
if (forwardReturn) {
iterator.return();
}
}
}
```
> The above code also has the additional nice property that it call `.return()`
> on the iterator when `n` values have been taken out of it.
That’s not what all the other constructs in ES6 do: they only call `return()`
if iteration stops abruptly.
Also missing from this code: checking whether the iterator actually has the
methods `return()` and `throw()` and responding accordingly.
--
Dr. Axel Rauschmayer
[email protected]
rauschma.de
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