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showing results for - "es6 reduce return promise"
Linus
23 Aug 2016
1const sum = await [
2  Promise.resolve(1),
3  Promise.resolve(1),
4  Promise.resolve(1)
5].reduce(async (previousPromise, itemPromise) => {
6  const sum = await previousPromise;
7  const item = await itemPromise;
8  return sum + item;
9}, Promise.resolve(0))
10
11// sum === 3
12
source
Sofia
25 Nov 2019
1async function slowCounter () {
2    const ret = await ([100, 200, 300, 400, 500]).reduce(
3        async (promise, wait, idx) => {
4            return promise.then(async last => {
5                const ret = last + wait
6                console.log(`${idx}: waiting ${wait}ms to return ${ret}`)
7                await new Promise((res, rej) => setTimeout(res, wait))
8                return ret
9            })
10        }, Promise.resolve(0))
11    console.log(ret)
12}
13
14slowCounter ()
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