async/await 中的递归未在主函数中解析

Question

我正在尝试在 asynс/await 函数中使用递归。问题是我无法在 main 函数中获得最终的承诺，其中我最初调用了递归方法

async function delay(ms) {
  return await new Promise(resolve => setTimeout(resolve, ms))
}

async function recursion(i) {
  return new Promise(async (resolve, reject) => {
    if (i == 0) {
      console.log(`i == 0`)
      resolve(i)
    } else {
      console.log(`i = ${i}. Wait 1 second...`)
      i--

      await delay(2000)
      await recursion(i)
    }
  })
}

async function main() {
  let i = await recursion(3)
  console.log(`END OF RECURSION`) //This code never use!
  console.log(`i => ${i}`)
}
main()

console.log:

i = 3. Wait 1 second...
i = 2. Wait 1 second...
i = 1. Wait 1 second...
i == 0

Answer 1

这是 promise 构造反模式。已经存在链式承诺，无需创建新承诺。从 Promise 回调返回的承诺将被忽略，这会破坏承诺链。此外，存在不一致，值并不总是从 recursion.

返回

应该是：

async function recursion(i) {
    if (i == 0) {
      console.log(`i == 0`)
      return i
    } else {
      console.log(`i = ${i}. Wait 1 second...`)
      i--

      await delay(2000)
      return recursion(i)
    }
}

async/await 中的递归未在主函数中解析

Recursion in async/await not resolve in main function

node.js

async-await

es6-promise