Javascript 变量作为对象指针
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Javascript Variables as Object Pointers
提问by nemo
I have a question out of curiosity. So I looked into how JS handles variable assignment and I get it. How does variable assignment work in JavaScript?
出于好奇，我有一个问题。所以我研究了 JS 如何处理变量赋值，我明白了。JavaScript 中的变量赋值是如何工作的？
But the same principle doesn't seem to exhibit itself in the following code I am working on:
但同样的原则似乎并没有在我正在处理的以下代码中体现出来：
var temp = playlist1[0];
playlist1[0] = playlist1[1];
playlist1[1] = temp;
I know this is a standard way to swap array elements. But if temp is pointing at playlist1[0], and playlist1[0]'s contents are changed to playlist1[1]'s then how come I don't end up with two playlist1[1]values in a row?
我知道这是交换数组元素的标准方法。但是如果 temp 指向playlist1[0]，并且playlist1[0]'s 的内容更改为playlist1[1]'s 那么为什么我不会playlist1[1]连续得到两个值？
回答by newacct
Not only variablesare object pointers. All values(that are not primitives) are object pointers. So tempis an object pointer. playlist1is a object pointer to an array object whose elements are object pointers. e.g. playlist1[0]is an object pointer, playlist1[1]is an object pointer, etc.
不仅变量是对象指针。所有值（不是基元）都是对象指针。temp对象指针也是如此。playlist1是指向数组对象的对象指针，其元素是对象指针。例如playlist1[0]是一个对象指针，playlist1[1]是一个对象指针，等等。
But if temp is pointing at playlist1[0]
但是如果 temp 指向 playlist1[0]
This doesn't make sense. tempis an object pointer. It points to an object. playlist1[0]is not an object; it's an object pointer. temp = playlist1[0];makes the object pointer temppoint to the same object as object pointer playlist1[0].
这没有意义。temp是一个对象指针。它指向一个对象。playlist1[0]不是一个对象；它是一个对象指针。temp = playlist1[0];使对象指针temp指向与对象指针相同的对象playlist1[0]。
If you know C, it is equivalent to something like this:
如果你知道 C，它就相当于这样：
Object *playlist1[10];
Object *temp = playlist1[0];
playlist1[0] = playlist1[1];
playlist1[1] = temp;
回答by TGH
This is consistent with the answer in the referenced question: You are just changing which object the variable points to - not the data it used to point to. Meaning temp is unaffected by the move to have playlist1[1] point to playlist1[2]. Temp retains the original value it pointed to when playlis1[1] and temp both pointed to it. Only playlist1[1] is updated
这与引用问题中的答案一致：您只是在更改变量指向的对象 - 而不是它曾经指向的数据。这意味着 temp 不受让 playlist1[1] 指向 playlist1[2] 的移动的影响。当 playlis1[1] 和 temp 都指向它时，Temp 保留它所指向的原始值。仅更新播放列表 1[1]
回答by mckurt
Because those are still references to elements in the array and not the elements themselves. So in the line:
因为这些仍然是对数组中元素的引用，而不是元素本身。所以在这一行：
playlist[1]=playlist[2]
You are not changing anything about temp. Contrast that with something like (assuming array elements were objects):
你没有改变任何关于温度的东西。对比一下（假设数组元素是对象）：
playlist[1].x=playlist[2].x
That is actually assigning the value of the object in the array, and if temp pointed to playlist[1], then temp.x would equal playlist[2].x
那实际上是在数组中分配对象的值，如果 temp 指向 playlist[1]，那么 temp.x 将等于 playlist[2].x
回答by bortunac
say we have obj={l1:{l2:[1,2]},}and we want to address obj.l1.l2[1]using an array of levels like arr=["l1","l2",1]then :
假设我们有 obj={l1:{l2:[1,2]},}并且我们想obj.l1.l2[1]使用像arr=["l1","l2",1]then这样的一组级别来解决：
Object.defineProperty(Object.prototype,'point',{
value:function(arr){
var rez=this;
for(var s in arr){
rez=rez[arr[s]];
if(rez === undefined) return undefined;
}
return rez;
}
});
So after defining "point" method (which is not enumerable to mess up everithing) we can use
因此，在定义“点”方法（不可枚举以弄乱一切）之后，我们可以使用
obj.point(arr)
to get value 2
获得价值 2
