How to transform an asynchronous function into a synchronous function in javascript?

Asked Nov 27 '19 at 13:56

Active Nov 27 '19 at 13:56

Viewed 51 times

I have an async function

const thirdPartyFoo = async () => { ... }

What I want to do is transform this function into a synchronous function, so when I call it, it awaits to be resolved to continue.

I can't use async/await or .then because here is my case.

I have a function foo that uses the thirdPartyFoo and it's called by other places in the application. I can't use .then because I need to return a value from thirdPartyFoo, but it can't be a promise.

function foo(){
    // some code

    thirdPartyFoo()

    // some code

   return bar // this can't be a Promise
}

What I need is something similar to the ajax async: false behavior.

How can I achieve this behavior giving any async function?

asked Nov 27 '19 at 13:56

Vencovsky

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1

In a nutshell: you can't. End of story. If `foo` calls an async function, it itself must be async, and anything that calls it must be too. Turtles all the way down. – deceze Nov 27 '19 at 14:00
2

`return Promise.reject("sad promise noises")` – Vencovsky Nov 27 '19 at 14:02
3

`.then(you => need.to('find', a => way['to']('deal')).with('it'))` – deceze Nov 27 '19 at 14:04
@deceze if I can't, how does `async: false` of the `ajax` do it? Going to make a question about it – Vencovsky Nov 28 '19 at 11:47

How to transform an asynchronous function into a synchronous function in javascript?

0 Answers0