I know that doing this:
const resultA = await a()
const resultB = await b()
// code here
Is effectively
a().then( resultA => {
b().then( resultB => {
// code here
})
})
Basically, a() runs then b() runs. I nested them to show that both resultA and resultB are in our scope; yet both function didn't run at once.
But what about this:
const obj = {
result1: await a(),
result2: await b()
}
do a() and b() run concurrently?
For reference:
const asyncFunc = async (func) => await func.call()
const results = [funcA,funcB].map( asyncFunc )
I know here funcA
and funcB
do run concurrently.
Bonus:
How would you represent the object assignment
const obj = {
result1: await a(),
result2: await b()
}
using then
/ callbacks?
UPDATE:
@Bergi is correct in this answer, this occurs sequentially. To share a nice solution for having this work concurrently for an object without having to piece together the object from an array, one can also use Bluebird
as follows
const obj2 = Bluebird.props(obj)
No, every await
will stop the execution until the promise has fulfilled, even mid-expression. It doesn't matter whether they happen to be part of the same statement or not.
If you want to run them in parallel, and wait only once for their result, you have to use await Promise.all(…)
. In your case you'd write
const [result1, result2] = await Promise.all([a(), b()]);
const obj = {result1, result2};
How would you represent the object assignment using
then
/ callbacks?
With temporary variables for each awaited value. Every await
translates into one then
call:
a().then(tmp1 => {
return b().then(tmp2 => {
const obj = {
result1: tmp1,
result2: tmp2
};
return …
});
})
If we wanted to be pedantic, we'd have to pick apart the object creation:
const tmp0 = {};
a().then(tmp1 => {
tmp0.result1 = tmp1;
return b().then(tmp2 => {
tmp0.result2 = tmp2;
const obj = tmp0;
return …
});
})