javascript - 循环对象并添加类型

Question

function arrayToList(array) {
  var list = null;
  for (var i = 0; i < array.length ; i++)
    list = {value: array[i], rest: list};
  return list;

}

console.log(arrayToList([10,20]));

为什么当我 console.log 这个数组时，我获取的是 value : 20、rest: object，而不是 值:10和其余:null。这让我很困惑，因为传入的数组的第一个元素是 10，而我将从 i = 0 开始。任何见解都会有帮助。谢谢。

Answer 1

这是因为您要返回最后一个设置为 list 的对象。要返回链表中的第一个，请在循环之前保留对其的引用，然后返回该引用。

虽然你的链接列表与数组相反。不确定这是否是您想要的。

如果您想列出镜像数组并转发链接，请执行以下操作:

function arrayToList(array) {
  var first = {value: array[0], rest: null};
  var prev = first;
  for (var i = 1; i < array.length ; i++)
    prev.rest = prev = {value: array[i], rest: null};

  return first;
}

console.log(arrayToList([10,20]));

<小时/>

假设我们没有循环，只是内联整个事情。这就是它的样子。我也将按照数组 [10, 20, 30] 的方式进行操作。

function arrayToList(array) {
  // Here `first` and `prev` reference the same object
  var first = {value: array[0], rest: null};
  var prev = first;

  // first = {value: 10, rest: null};
  // prev = {value: 10, rest: null};

  // Then we assign a new object to `prev.rest` as well as `prev`.
  // Because the `prev` and `first` variables reference the same object,
  // the mutation of `prev.rest` is also seen from `first.rest`.

  // First iteration
  prev.rest = prev = {value: array[1], rest: null};

  // now...
  // first = {value: 10, rest: (reference to object with value: 20)};
  // prev = {value: 20, rest: null};

  // So it first changed `prev.rest` (and therefore `first.rest`) to point to
  // the new object, but then it also updated `prev` to the same, new object.

  // Take note that though we assigned the new object to `prev.rest`, after
  // the assignment, `prev.rest` is actually `null`, because the `prev.rest`
  // that we assigned to was the `.rest` of the *previous* `prev`, which was
  // immediately updated to see the *new* object.

  // Second iteration
  prev.rest = prev = {value: array[2], rest: null};

  // now...
  // first = {value: 10, rest: (reference to object with value: 20)};
  // prev = {value: 30, rest: null};

  // So now when we changed `prev.rest`, it had no impact on `first` because
  // `prev` and `first` are referencing different objects. So `prev` sees the
  // new object, but `first` does not.

  return first;
}