javascript - 在原型上实现映射时加深对回调方法的了解(Increasing understanding of callback method when implementing map on a Prototype)

Question

var s = [23, 65, 98, 5];

Array.prototype.myMap = function (callback) {
  var newArray = []; 
  this.forEach(function (a) {
    return newArray.push(callback(a));
  }); 
  return newArray;
};

var new_s = s.myMap(function (item) {
  return item * 2;
});

myMap is a function we have defined that has a param called callback inside myMap is a forEach that executes a function once on each array element.

(myMap是一个我们定义的函数，在myMap中有一个名为callback的参数， myMap是一个forEach ， myMap每个数组元素执行一次函数。)

It's this line I don't understand

(我不明白这是这条线)

return newArray.push(callback(a));

so push() to newArray

(所以push()到newArray)

MDN push() : - https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/push

(MDN push（） ： -https : //developer.mozilla.org/zh-CN/docs/Web/JavaScript/Reference/Global_Objects/Array/push)

But push() accepts only one parameter.

(但是push()仅接受一个参数。)

So why do we have .push(callback(a)) why not just .push(a) ?

(那么为什么我们有.push(callback(a))为什么不只是.push(a) ？)

  ask by user1554264 translate from so

Answer 1

But push() accepts only one parameter.

(但是push（）仅接受一个参数。)

So why do we have .push(callback(a))

(那么为什么会有.push（callback（a））)

The return value of callback(a) is only one parameter.

(的返回值callback(a) 只有一个参数。)

why not just .push(a)?

(为什么不只是.push（a）？)

The point of mapping is to take an array of data and create a new array where each item is transformed in some way.

(映射的重点是获取一个数据数组并创建一个新数组，其中每个项目都以某种方式进行转换 。)

If you didn't pass a through the callback function, it wouldn't be transformed, and you'd just be making a shallow copy.

(如果您没有通过回调函数传递a ，则该函数将不会被转换，而您只是制作一个浅表副本。)