javascript - 通过引用和拼接传递

Question

我正在尝试更好地理解引用传递如何与拼接相关。我想要一个主对象数组，并使用另一个引用主数组对象的数组。例如:

var mainArray = [
    someObject,
    anotherObject
]
var subArray = [];
subArray.push( mainArray[ 0 ] );

我不明白的是如果我拼接原始值会发生什么？

mainArray.splice( 0, 1 );

是否子数组[0]:

1. 也被拼接了？
2. 值仍然存在，因为有指向它的指针？
3. 留下一个空值？

Answer 1

在这段代码之后:

var mainArray = [
    someObject,
    anotherObject
];
var subArray = [];
subArray.push( mainArray[ 0 ] );

...你在内存中的内容看起来有点像这样:

                                                   +−−−−−−−−−−+someObject−−−−−−−−−−−−−−−−−−+−−+−−−−−−−−−−−−−−−−−−>| (object) |                           /  /                    +−−−−−−−−−−+                           |  |                               |  |                     +−−−−−−−−−−+                           |  |  anotherObject−−−+−>| (object) |                           |  |                 /   +−−−−−−−−−−+                           |  |                |             +−−−−−−−−−−−+ |  |                |subArray−−−−>| (array)   | |  |                |             +−−−−−−−−−−−+ |  |                |             | length: 1 | |  |                |             | 0:        |−/  |                |             +−−−−−−−−−−−+    |                |                              |                |              +−−−−−−−−−−−+   |                |mainArray−−−−>| (array)   |   |                |              +−−−−−−−−−−−+   |                |              | length: 2 |   |                |              | 0:        |−−/                 |              | 1:        |−−−−−−−−−−−−−−−−−−−−/              +−−−−−−−−−−−+

Note that there is no connection between mainArray and subArray at all. They both just happen to contain the same value (which is also the value in someObject, all three places have an object reference for the same object).

Removing that value from mainArray has no effect on subArray (or on the objects):

mainArray.splice(0, 1);

                                                   +−−−−−−−−−−+someObject−−−−−−−−−−−−−−−−−−+−−−−−−−−−−−−−−−−−−−−−>| (object) |                           /                       +−−−−−−−−−−+                           |                                  |                        +−−−−−−−−−−+                           |     anotherObject−−−+−>| (object) |                           |                    /   +−−−−−−−−−−+                           |                   |             +−−−−−−−−−−−+ |                   |subArray−−−−>|  (array)  | |                   |             +−−−−−−−−−−−+ |                   |             | length: 1 | |                   |             | 0:        |−/                   |             +−−−−−−−−−−−+                     |                                               |              +−−−−−−−−−−−+                    |mainArray−−−−>|  (array)  |                    |              +−−−−−−−−−−−+                    |              | length: 1 |                    |              | 0:        |−−−−−−−−−−−−−−−−−−−−/              +−−−−−−−−−−−+

因此基于此:

1. also get spliced?

不，改变 mainArary 的内容对 subArray 没有影响。

2. the value still exist since there is a pointer to it?

subArray 条目 0 的值不受影响。

3. get left with an empty value?

不，对 subArray 根本没有影响。

---

I am trying to better understand how pass by reference works in relation to splice.

您混淆了“引用”一词的两种不同含义(很多人都这样做)。

“按引用传递”是编程中的一个术语。也就是说，它具有特定的含义:将对变量 的引用传递给函数，这样函数就可以访问并更改该变量的内容。 JavaScript 没有传递引用。

你在问题​​中处理的是对象引用，它与传递引用无关，只是这两个概念都包含“引用”一词(但在关系完全不同的东西。)

对象引用是值。它们告诉 JavaScript 引擎对象在内存中的位置。