异步/等待中的递归不能在main函数中解析

Question

我正在尝试使用asynс/ await函数中的递归。问题是我无法在main函数中得到最终的承诺，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()

的console.log：

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

Answer 1

这是承诺建筑反模式。已经存在链条的承诺，无需创建新的链条。从Promise回调返回的承诺被忽略，这打破了承诺链。此外，存在不一致性，该值并非总是从recursion返回。

它应该是：

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)
    }
}