关于javascript:js中pushpopslicemap-和-reduce实现

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;
  }