不要忘记js是单线程的，所以主线程上的计算是以牺牲其他函数为代价的，所以例如p1和p2“并行”运行，但在同一个事件循环上。

const p1 = async () => {
  let x = 0
  while(x < 100000000) {x++}
  return x
}
const p2 = async () => {
  let y = 100000000
  while(y > 0) {y--}
  return y
}
const p3 = async () => {
  let z = 0
  while(z < 200000000) {z = z + 2}
  return z
}

const run1 = async () => {
   console.time("Parallel")
   const [res1, res2, res3] = await Promise.all([p1(), p2(), p3()])
   console.timeEnd("Parallel")
}
const run2 = async () => {
   console.time("Sync")
   const res1 = await p1()
   const res2 = await p2()
   const res3 = await p3()
   console.timeEnd("Sync")
}

const p4 = async () => {
  return new Promise(resolve => {
    timeout = setTimeout(() =>  {
      resolve(true);
    }, 100);
  });
}
const p5 = async () => {
  return new Promise(resolve => {
    timeout = setTimeout(() =>  {
      resolve(true);
    }, 150);
  });
}
const p6 = async () => {
  return new Promise(resolve => {
    timeout = setTimeout(() =>  {
      resolve(true);
    }, 200);
  });
}

const run3 = async () => {
   console.time("Parallel2")
   const [res1, res2, res3] = await Promise.all([p4(), p5(), p6()])
   console.timeEnd("Parallel2")
}
const run4 = async () => {
   console.time("Sync2")
   const res1 = await p4()
   const res2 = await p5()
   const res3 = await p6()
   console.timeEnd("Sync2")
}

(async()=>{
  await run1()
  await run2()
  await run3()
  await run4()
})()

正如你在第二个例子中看到的，对于非计算密集型任务，Promise.all需要的时间与它最长的promise一样长，而链接则需要所有promise的总和。