昂贵的计算示例

时间:2020-06-27 07:32:37

标签: javascript performance

我正在学习如何衡量性能,并且想使用一些大约需要1-2秒的代码来进行计算。

我使用了下面摘自Mozilla的这段代码,但我想知道是否有人可以帮助我进行更简洁的操作。不必“有意义”。

const iterations = 50;
const multiplier = 1000000000;

function calculatePrimes(iterations, multiplier) {
  var primes = [];
  for (var i = 0; i < iterations; i++) {
    var candidate = i * (multiplier * Math.random());
    var isPrime = true;
    for (var c = 2; c <= Math.sqrt(candidate); ++c) {
      if (candidate % c === 0) {
        // not prime
        isPrime = false;
        break;
      }
    }
    if (isPrime) {
      primes.push(candidate);
    }
  }
  return primes;
}

function doPointlessComputationsWithBlocking() {
  var primes = calculatePrimes(iterations, multiplier);
  pointlessComputationsButton.disabled = false;
  console.log(primes);
}

1 个答案:

答案 0 :(得分:1)

以“需要时间”为目的进行操作的最简单方法是只在一段时间内什么也不做:

function wait(seconds) {
  var start = new Date();
  //empty while loop until the required amount of time has passed
  while((new Date() - start) / 1000 < seconds);
}

var begin = performance.now();

console.log("start");

wait(2);

console.log("finish in", performance.now() - begin, "ms");

如果您不确定需要的时间,则可以随意增加或修改等待时间:

function wait(seconds) {
  // add or subtract up to 50% 
  seconds *= Math.random() + 0.5;
  var start = new Date();
  while((new Date() - start) / 1000 < seconds);
}

var begin = performance.now();

console.log("start");

wait(2);

console.log("finish in", performance.now() - begin, "ms");