push办法

Array.prototype.push = function (...items) {    let O = Object(this); // ecma 中提到的先转换为对象    // >>> 无符号右移    let len = this.length >>> 0    let argCount = items.length >>> 0    // 2 ^ 53 - 1 为JS能示意的最大正整数    if (len + argCount > 2 ** 53 - 1) {        throw new TypeError("The number of array is over the max value")    }    for (let i = 0; i < argCount; i++) {        O[len + i] = items[i];    }    let newLength = len + argCount;    O.length = newLength;    return newLength;}

pop 办法

Array.prototype.pop = function() {  let O = Object(this);  let len = this.length >>> 0;  if (len === 0) {    O.length = 0;    return undefined;  }  len --;  let value = O[len];  delete O[len];  O.length = len;  return value;}

map办法

Array.prototype.map = function(callbackFn, thisArg) {  if (this === null || this === undefined) {    throw new TypeError("Cannot read property 'map' of null");  }  if (Object.prototype.toString.call(callbackfn) != "[object Function]") {    throw new TypeError(callbackfn + ' is not a function')  }  let O = Object(this);  let T = thisArg;  let len = O.length >>> 0;  let A = new Array(len);  // 赋值给新数组，并不扭转原数组的值  for(let k = 0; k < len; k++) {    if (k in O) {      let kValue = O[k];      // 顺次传入this, 以后项，以后索引，整个数组      let mappedValue = callbackfn.call(T, KValue, k, O);      A[k] = mappedValue;    }  }  return A;}

reduce办法
两个留神点

  • 初始值默认值不传的非凡解决；

  • 累加器以及 callbackfn 的解决逻辑。

    Array.prototype.reduce  = function(callbackfn, initialValue) {// 异样解决，和 map 相似if (this === null || this === undefined) {  throw new TypeError("Cannot read property 'reduce' of null");}// 解决回调类型异样if (Object.prototype.toString.call(callbackfn) != "[object Function]") {  throw new TypeError(callbackfn + ' is not a function')}let O = Object(this);let len = O.length >>> 0;let k = 0;let accumulator = initialValue;  // reduce办法第二个参数作为累加器的初始值if (accumulator === undefined) {      throw new Error('Each element of the array is empty');    // 初始值不传的解决  for(; k < len ; k++) {    if (k in O) {      accumulator = O[k];      k++;      break;    }  }}for(;k < len; k++) {  if (k in O) {    // 留神 reduce 的外围累加器    accumulator = callbackfn.call(undefined, accumulator, O[k], O);  }}return accumulator;}