Is the code below guaranteed to output HERE?

var p = new Promise(() => console.log("HERE"))

(That is, does var p = new Promise(fn) always execute fn if p.then(…) is never called to do something with the result?)

More specifically, in the context of service workers, if I call Cache.delete() but never call .then() on the return value (or I throw away the return value), is the cache entry guaranteed to be deleted?

Is the code below guaranteed to output HERE?

var p = new Promise(() => console.log("HERE"))

(That is, does var p = new Promise(fn) always execute fn if p.then(…) is never called to do something with the result?)

More specifically, in the context of service workers, if I call Cache.delete() but never call .then() on the return value (or I throw away the return value), is the cache entry guaranteed to be deleted?

• javascript
• promise
• lazy-evaluation
• service-worker

Share Improve this question Follow asked Feb 3, 2016 at 12:32 mjsmjs 65.4k2727 gold badges9696 silver badges129129 bronze badges

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2 Answers 2

Sorted by: Reset to default 6

Yes, it is guaranteed. The specification of Promise has this step which will always be evaluated:

9. Let pletion be Call(executor, undefined, «resolvingFunctions.[[Resolve]], resolvingFunctions.[[Reject]]»).

where executor is what you passed to the Promise constructor, and Call results in that code being run. This all happens before the Promise is even returned to your p variable.

As James said, it is guaranteed that the function will be called. Though this doesn't guarantee that the cache entry gets deleted!

You have to check the value of the promise resolution (true if the cache entry is deleted, false otherwise).