Re: How would shallow generators compose with lambda?

> To make this clear with an example (thanks to Jason for some IRC
> interaction):
>
> function gen(arg) {
>     foo((lambda (x) yield x), arg);
> }
> function foo(callback, arg) {
>     try {
>         callback(arg);
>     } finally {
>         alert("I'm ok!");
>     }
> }
> g = gen(42);
> print(g.next()); // tell the user the meaning of life, etc.
> g = null;
> gc();
>
> I think finally is the only issue, since how else can you tell that foo
> didn't see a return or exception from the callback?
>
> It's true that this finally-may-not-run wrinkle is confined to generator
> functions in JS1.7, i.e., without lambda.
>
> But I think any function that invokes an arbitrary other function in a
> try has to fear that its finally might not run. There could be a hard
> stop due to iloop prevention, or an uncatchable security fail-stop error.
>
> FWIW I'm cool on lambda and return to label atm.
>
> /be