javascript await in the caller and in the callee

Question

I have a service class called: Test

That Test Class has a function called: getMe()

In that getMe function, I have 3 await statements, 2nd needs the answer from the first one, third one needs an answer from the second one. An example:

class Test {
   async getMe() {
      const res1 = await promise1;
      const res2 = await promise2(res1);
      const res3 = await promise3(res2);
      return res3;
   }
}

Now, somewhere in my code, i call this function.

const a = new Test();
try{
  const res = await a.getMe();
}catch(err){
  console.log("error", err);
}

Now, because of the fact that in getMe function, I already await promises, It's not optimized since there's a middle promise created. Read this: Difference between `return await promise` and `return promise`

So, I am wondering, If in the getMe function, I shouldn't write await for optimizations and return promise directly, how could my code be written ? I don't want in my outside code to call await promise1, promise2, promise3 because then I'd not have a single function which takes care of the final bits and my code would be scattered.

What do you think and could be your advice ?

eol · Accepted Answer · 2020-07-08T11:36:33.827

Since the promises inside getMe() are dependent on each other (e.g. promise2 requires the resolved value of promise1 as an input-parameter), you cannot really optimize something here (apart from directly returning the last promise).

If they weren't dependent on other you could certainly use Promise.all() in order to process them non-sequentially, return its promise and just await this promise in the calling function.

score 1 · Answer 2 · answered Jul 08 '20 at 11:35

The only thing you could do to optimize it, would be to return the result of promise3 directly.

class Test {
   async getMe() {
      const res1 = await promise1;
      const res2 = await promise2(res1);
      return promise3(res2);
   }
}

Other than that, I dont see any way, as the three functions are dependent on one another.

javascript await in the caller and in the callee

2 Answers2