Javascript背景循环

gra*_*tot 5 javascript asynchronous event-loop node.js background-task

假设我们有一个loop.js文件:

longLoop().then(res => console.log('loop result processing started'))
console.log('read file started')
require('fs').readFile(__filename, () => console.log('file processing started'))
setTimeout(() => console.log('timer fires'), 500)

async function longLoop () {
  console.log('loop started')
  let res = 0
  for (let i = 0; i < 1e7; i++) {
    res += Math.sin(i) // arbitrary computation heavy operation
    if (i % 1e5 === 0) await null /* solution: await new Promise(resolve => setImmediate(resolve)) */
  }
  console.log('loop finished')
  return res
}
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如果ran(node loop.js)输出:

loop started
read file started
loop finished
loop result processing started
timer fires
file processing started
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当循环在后台运行时,如何重写此代码以读取和处理文件?

我的解决方案

我想出的是:

longLoop().then(res => console.log('loop result processing started'))
console.log('read file started')
require('fs').readFile(__filename, () => console.log('file processing started'))
setTimeout(() => console.log('timer fires'), 500)

async function longLoop () {
  let res = 0
  let from = 0
  let step = 1e5
  let numIterations = 1e7
  function doIterations() {
    //console.log(from)
    return new Promise(resolve => {
      setImmediate(() => { // or setTimeout
        for (let i = from; (i < from + step) && (i < numIterations); i++) {
          res += Math.sin(i)
        }
        resolve()
      })
    })
  }
  console.log('loop started')
  while (from < numIterations) {
    await doIterations()
    from += step
  }
  console.log('loop finished')
  return res
}
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确实记录了:

loop started
read file started
file processing started
timer fires
loop finished
loop result processing started
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有更简单,更简洁的方法吗?我的解决方案有什么缺点?

tri*_*cot 3

代码的第一个版本阻止进一步处理的原因是await获得立即解决的承诺(值null被包装在承诺中,就像您所做的那样await Promise.resolve(null))。这意味着之后的代码将在当前await“任务”期间恢复:它只是将微任务推送到任务队列中,该微任务将在同一任务中被消耗。您待处理的所有其他异步内容都在任务队列中等待,而不是微任务队列。

对于 来说是这样setTimeout,对于 也是这样readFile。它们的回调在任务队列中待处理,因此不会获得比 s 生成的微任务更高的优先级await

因此,您需要一种方法来将await某些内容放入任务队列而不是微任务队列中。您可以通过向其提供一个承诺来做到这一点,该承诺不会立即解决,而只会在当前任务之后解决。

你可以用 .... 引入延迟setTimeout

const slowResolve = val => new Promise(resolve => setTimeout(resolve.bind(null, val), 0));
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您可以使用await. 这是一个使用图像加载而不是文件加载的片段,但原理是相同的:

const slowResolve = val => new Promise(resolve => setTimeout(resolve.bind(null, val), 0));
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const slowResolve = val => new Promise(resolve => setTimeout(resolve.bind(null, val), 0));

longLoop().then(res => 
    console.log('loop result processing started'))

console.log('read file started')

fs.onload = () => 
    console.log('file processing started');
fs.src = "https://images.pexels.com/photos/34950/pexels-photo.jpg?h=350&auto=compress&cs=tinysrgb";

setTimeout(() => console.log('timer fires'), 500)

async function longLoop () {
  console.log('loop started')
  let res = 0
  for (let i = 0; i < 1e7; i++) {
    res += Math.sin(i) // arbitrary computation heavy operation
    if (i % 1e5 === 0) await slowResolve(i);
  }
  console.log('loop finished')
  return res
}
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