Re: [fetch] Aborting a fetch (#27)

> should silently keep going because there was nothing to cancel ... right ?

Yes, since all the promises in the array are `fulfilled` the promise returned by `Promise.all()` will also be fulfilled, so it is too late to cancel it. Maybe the `cancel()` could return a `pending` promise:
```js
var p = Promise.all([Promise.resolve()]);
//p.@state = "fulfilled" 
var q = p.cancel();
//p.@state = "fulfilled"
//q.@state = "pending"
```

> if a cancelable Promise returns inside a then another cancelable Promise, should the last cancel eventually cancel this one, instead of the original one that generated the chain?

The promise returned by `then(func)` is not the promise returned by the function `func`:
```js
var innerP;
var outerP = Promise.resolve().then(r => (innerP = new Promise(r => r()));
outerP === innerP //false
```
When calling `c = a.then(() => b)`, then `c` is a promise that is only ever waiting for one other promise to resolve, either the previous one in the chain (`a`) or the promise returned by the function passed to it (`b`). This means that calling `cancel` on `c` will only cancel the one it is waiting for. If `a` hasn't resolved yet, then `a` will be cancelled and nothing will happen to `b`, since it hasn't been started yet. If `a` has resolved, then `a` will not be cancelled (since it is resolved), and `b` will be cancelled. If `b` has resolved, then `c` has resolved, and cancelling `c` will be a no-op. It will be too late.



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