javascript中promise的执行顺序

Question

I have a sample program to understand how promises and async/ await works. But am a little bit confused about the execution order of promises. Can anybody explain how this works?

CASE1

//Promise 1
let promiseTest = new Promise((resolve, reject) => {
  var k = 0;
  for(i=0; i< 1000; i++ ){
    k += i;
  }
  resolve(k);
  console.log("Inside promise1")
});

promiseTest.then((res)=> {
    console.log('Promise1 result : '+ res);
}).then(() => {
    promiseTest2.then((res) => {
        console.log(res)
    });
}).then(finish)
.catch((err) => {
 console.log(err)
});

//Promise 2
let promiseTest2 = new Promise ((resolve, reject) => {
    console.log("Inside promise2")
});

function finish(){
    console.log("finished promise");
}

For this am geting the result as

RESULT

Inside promise1
Inside promise2
Promise1 result : 499500
finished promise

CASE2

I have another example doing the same with async/ await. But in this order of execution is correct.

//Async await test
async function AsyncTest(){
    console.log("Inside async1")
    var k = 0;
    for(i=0; i< 1000; i++ ){
      k += i;
    }
  console.log('async1 result : '+ k);
  const result =  await AsyncTest2();
  console.log(result)
  console.log("finished async");
}

async function AsyncTest2(){
   return "Inside async2";
}

AsyncTest();

RESULT

Inside async1
async1 result : 499500
Inside async2
finished async

Thanks.

Answer 1

If you call .then(cb) on a Promise, a new Promise gets created and returned by it, that will resolve to whatever the callbback returns . If that is a Promise itself, that Promise will be awaited before the chain continues. In your case you did:

promiseTest.then((res)=> {
 console.log('Promise1 result : '+ res);
 return undefined; // implicit return
}).then(() => {
 /* doesnt matter what you do here */
 return undefined;
}).then(finish)

Wether you attach another .then to another promise does not matter to this chain of promises.