Javascript:在异步函数中返回一个 promise

Question

如果我有，会有什么不同吗:

async function test () {
  const foo = await bar()
  return Promise.all([promise1, promise2])
}

代替:

async function test () {
  const foo = await bar()
  const [result1, result2] = await Promise.all([promise1, promise2])
  // Given that I don't care about result1, result2 in this `test` function
  return [result1, result2]
}

如果我这样做，我会得到相同的结果。例如。对于任何一种情况，我都可以这样做:

test().then(([result1, result2]) => { ... })

但我更好奇它们如何表现相同的底层机制。

换句话说，如果我在函数内部返回一个 promise 而不是一个值，异步函数如何处理它？<​​/p>

Answer 1

我认为您在 promise 链中使用 await 有效地调用类似同步的函数，根据此 answer :

You are perfectly free to call either synchronous functions within the promise chain (from within .then() handlers) or asynchronous functions that then return a new promise.

When you return something from a .then() handler, you can return either a value (which becomes the resolved value of the parent promise) or you can return another promise (which chains onto the previous promise) or you can throw which works like returning a rejected promise (the promise chain becomes rejected).