为什么 Javascript 不“等待”等待正确的时间？

ceremcem 发表于 Dev

陶瓷

此代码无法按预期工作：

function sleep(ms) {
  return new Promise(resolve => setTimeout(resolve, ms));
}

function diff(expected) {
  let x = expected - Date.now();
  if (x > 0) {
    return `earlier ${x} ms`;
  } else if (x < 0) {
    return `late ${-x} ms`;
  } else {
    return `just in time`;
  }
}

start = Date.now();
async function demo() {
  let loop_count = 5;
  for (let i = 0; i < loop_count; i++) {
    console.log(diff(start + i * 1000) + `: Waited ${i} seconds...`);
    await sleep(i * 1000);
  }
  console.log(diff(start + loop_count * 1000) + ': Done');
}

demo();

输出是：

$ node test.js 
just in time: Waited 0 seconds...
earlier 993 ms: Waited 1 seconds...
earlier 993 ms: Waited 2 seconds...
late 10 ms: Waited 3 seconds...
late 2011 ms: Waited 4 seconds...
late 5013 ms: Done

这种差异从何而来？

t.niese

这种差异从何而来？

因为你测量不正确。首先你console.log(diff(start + i * 1000) + `: Waited ${i} seconds...`);在await sleep(i * 1000);. 此测试必须在sleep.

另一个问题是您不会更改该start值，并且您始终使用与start每次迭代的参考相同的值。

所以在i>1你得到错误的结果之后：你已经等了1第二个，所以 for i=2the total time elapsed from startis (1 + 2)seconds, for i=3to total time elapsed from startis (1 + 2 + 3)seconds, ... 。

更新代码以使之前的睡眠时间也包含在计算中，从而得到预期的结果：

function sleep(ms) {
  return new Promise(resolve => setTimeout(resolve, ms));
}

function diff(expected) {
  let x = expected - Date.now();
  if (x > 0) {
    return `earlier ${x} ms`;
  } else if (x < 0) {
    return `late ${-x} ms`;
  } else {
    return `just in time`;
  }
}

function sumUp(num) {
  let res = 0;
  for(let i=0; i<=num ; i++) {
    res += i;
  }
  return res;
}

start = Date.now();
async function demo() {
  let loop_count = 5;
  for (let i = 0; i < loop_count; i++) {
    await sleep(i * 1000);
    console.log(diff(start + sumUp(i) * 1000) + `: Waited ${i} seconds...`);
  }
  console.log(diff(start + sumUp(loop_count) * 1000) + ': Done');
}

demo();