为什么splice()方法会更改其他数组的值?

I have an array say

我有一个数组。

var list = ["first", "second"];

now I assign list to some other variable say

现在我把列表赋给另一个变量。

var temp = list;

Now when I use splice on list like

现在当我在列表中使用splice时

list.splice(0,1);

Now when I check the value of list it shows

现在，当我检查列表的值时，它会显示出来

list = ["second"]

Also when I check the value of temp then it says

当我检查temp的值时，它会说

temp = ["second"]

I want to know why is that so? Why the value of temp is changed?

我想知道为什么是这样?为什么要更改temp的值?

4 个解决方案

#1

when using var temp = list you're creating a reference temp to list. So since splice changes list array in-place, you're also changing temp

当使用var temp = list时，您正在创建要list的引用temp。因此，既然splice改变了列表数组，那么您也在改变temp。

Use slice instead which returns a copy of the array, like so

使用slice返回数组的一个拷贝，就像这样

var temp = list.slice();

#2

A common mistake in JS

JS中常见的错误

This is a pointer, not a clone of the object:

这是一个指针，不是对象的克隆:

var temp = list;

If you want to actually copy the object, there are a few ways. The simplest is just to concat it with itself:

如果你想复制这个对象，有几种方法。最简单的办法就是把它和它自己联系起来:

var temp = list.concat([]);
// or
var temp = list.slice();

Note that this is somewhat dangerous, it only gets the base values out of the array. There are more advanced methods for cloning objects and creating 'perfect' array clones.

注意，这有点危险，它只从数组中获取基值。还有更高级的方法可以克隆对象并创建“完美”的数组克隆。

#3

slice

片

should do the trick:

应该是可以奏效的:

var temp = list.slice(0);

var temp = list.slice(0);

About object cloning, look here How do you clone an Array of Objects in Javascript?

关于对象克隆，请看这里，如何在Javascript中克隆对象数组?

#4

You can use a prototype method to add this functionality.

您可以使用一个原型方法来添加这个功能。

Object.prototype.clone = function() {
  var newObj = (this instanceof Array) ? [] : {};
  for (i in this) {
    if (i == 'clone') continue;
    if (this[i] && typeof this[i] == "object") {
      newObj[i] = this[i].clone();
    } else newObj[i] = this[i]
  } return newObj;
};

Which then allows you to do this:

这样你就可以做到:

var a = [1,2,3];

var b = a.clone();

a[1] = 5;

console.log(a); //1,5,3
console.log(b); //1,2,3

Disclaimer: shamelessly borrowed from here (the whole article is worth a read): http://my.opera.com/GreyWyvern/blog/show.dml/1725165

免责声明:无耻地从这里借用(整篇文章值得一读):http://my.opera.com/GreyWyvern/blog/show.dml/1725165

#1