javascript - 异步函数中没有执行代码

Question

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Answer 1

不确定这是否是一个拼写错误，但您忘记调用 resolve 所以 await 之后的下一行将执行:

function setTimeoutSync(fn, milli) {
  return new Promise((resolve) => {
    setTimeout(() => {
      fn()
      resolve()
    }, milli);
  });
}

async function run() {
  console.log('start');
  await setTimeoutSync(() => {
    console.log('setTimeoutSync');
  }, 3000);
  console.log('end');
}

run();

关于你的问题:

Can you explain how not calling resolve can result in the rest of the function not executing?...

嗯async await使用 generators和 promise行为:

The purpose of async/await functions is to simplify the behavior of using promises synchronously and to perform some behavior on a group of Promises. Just as Promises are similar to structured callbacks, async/await is similar to combining generators and promises.

发电机可以 yield结果因此不会终止执行上下文，并且可以在它停止的地方继续执行。

基本上你的例子可以写成Promise.prototype.then()回调:

function setTimeoutSync(fn, milli) {
  return new Promise((resolve) => {
    setTimeout(() => {
      fn()
      resolve()
    }, milli);
  });
}

async function run() {
  console.log('start');
  setTimeoutSync(() => {
    console.log('setTimeoutSync');
  }, 3000)
  .then((result) => {
  	console.log('end');
  });
  
}

run();

如您所见，如果我们不解析，则 .then 的回调将不会运行。