​​题目来源：

        leetcode题目，网址：2725. 间隔取消 - 力扣（LeetCode）

解题思路：

        利用 setInterval（）每隔一段时间执行，利用闭包获取停止时间。 

解题代码：

/*** @param {Function} fn* @param {Array} args* @param {number} t* @return {Function}*/
var cancellable = function(fn, args, t) {fn(...args);const time=setInterval(()=>fn(...args),t);return ()=>clearInterval(time);
};/***  const result = []**  const fn = (x) => x * 2*  const args = [4], t = 35, cancelT = 190**  const start = performance.now()**  const log = (...argsArr) => {*      const diff = Math.floor(performance.now() - start)*      result.push({"time": diff, "returned": fn(...argsArr)})*  }*       *  const cancel = cancellable(log, args, t);**  setTimeout(() => {*     cancel()*  }, cancelT)*   *  setTimeout(() => {*    console.log(result)  // [*                         //      {"time":0,"returned":8},*                         //      {"time":35,"returned":8},*                         //      {"time":70,"returned":8},           *                         //      {"time":105,"returned":8},*                         //      {"time":140,"returned":8},*                         //      {"time":175,"returned":8}*                         //  ]*  }, cancelT + t + 15)    */

总结：

        抄的。        

        setTimeOut();  //在指定的毫秒数后调用函数或计算表达式

        setInterval();        //每隔一定时间调用函数、方法或对象