javascript - 对此行为 : ES6 Promise that is never resolved 的解释是什么

Question

以下代码:

Promise.resolve('a string')
  .then(resp => {
    console.log('resp from the initial promise: ' + resp)
    const p0 = new Promise((resolve, reject) => {
      setTimeout(() => {
        console.log('finished sleeping in p0')
        resolve('hello from p0')
      }, 5000)
    })
    console.log('returning p0')
    return p0
  })
  .then(resp => {
    console.log('resp from p0: ' + resp)
    const p1 = new Promise((resolve, reject) => {
      console.log('p1 completed')
      // resolve('another string from p1')
    })
    console.log('returning p1')
    return p1
  })
  .then(res => {
    console.log('resp from p1: ' + res)
  })
  .catch(err => {
    console.error('Error: ' + err.message)
  })

console.log('last statement')

当我运行它时，我得到以下输出(这对我来说没有意义)

last statement
resp from the initial promise: a string
returning p0
finished sleeping in p0
resp from p0: hello from p0
p1 completed
returning p1

首先，由于 promise p1 从未得到解决，我希望程序永远等待并且永远不会完成。事实并非如此，它完成得很好(尽管没有到达最后一个 then())。

此外，它还创建 promise 并在 创建 promise 之后的代码之前执行 promise 内的代码。我希望“返回 p1”在“p1 完成”之前出现，因为我假设 promise 中的内容将在下一个滴答时执行。

Answer 1

Also, it creates the promise and executes the code inside of the promise before the code that comes after it's created. I would expect 'returning p1' to come before 'p1 completed' since I assumed that what's inside the promise would be executed on the next tick.

如果您在一个 promise 上调用 .then，您传入的函数将延迟到下一个 tick。但是 promise 的构造函数不会那样做。当您构造一个 new Promise 时，您传递给构造函数的代码会立即同步运行。所以这段代码:

const p0 = new Promise((resolve, reject) => {
  setTimeout(() => {
    console.log('finished sleeping in p0')
    resolve('hello from p0')
  }, 5000)
})
console.log('returning p0')

... 将立即设置超时，甚至在 promise 被分配给 p0 之前。这段代码:

const p1 = new Promise((resolve, reject) => {
  console.log('p1 completed')
  // resolve('another string from p1')
})
console.log('returning p1')

... 将立即注销“p1 已完成”，甚至在 promise 被分配给 p1 之前。