为什么在 b 打印之前打印 2 和 4 打印? [英] why 2 and 4 prints before b prints?

问题描述

function first() {
  return new Promise(resolve => {
    console.log(2);
    resolve(3);
    console.log(4);
  });
}

async function f() {
  console.log(1);
  let r = await first();
  console.log(r);
  console.log(99);
}

console.log('a');
f();
console.log('b');

在上面的代码中显示如下结果:

In the above code shows the following result:

a
1
2
4
b
3
99

据我了解，当编译器遇到await first()函数时，会将first()函数的执行推入事件队列并暂停f()，继续执行f()之后的一切，所以执行顺序应该是:

In my understanding, when the compiler hits the await first() function, it pushes the first() function execution to the event queue and pause the execution of f(), continue execution everything after f().So the execution order should be:

a
1
b
2
4
3
99

显然，我理解错了.谁能向我解释这是如何真正起作用的?

Apparently, I get it wrong. Can anyone explain to me how this is really working?

推荐答案

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if you didn't understand ,leave a comment . Thank you.

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