如何理解#dup和#clone对引用其他对象的对象进行操作？

问题描述：

在 ruby 的文档中，我不确定 ...但不是它们引用的对象的含义 code> rubinus 。

I am not sure about the meaning of "...but not the objects they reference" in both the documantion of ruby and rubinus.

在红宝石，其中有 #clone 和的解释#dup 行为说：

产生obj的浅表副本-obj的实例变量为
复制，但不复制它们引用的对象。复制obj的冻结和
污染状态。另请参见Object＃dup下的讨论。

Produces a shallow copy of obj—the instance variables of obj are copied, but not the objects they reference. Copies the frozen and tainted state of obj. See also the discussion under Object#dup.

在鲁比尼乌斯：

复制实例变量，但不递归复制它们引用的对象
。

Copies instance variables, but does not recursively copy the objects they reference. Copies taintedness.

我尝试了以下代码，但是行为超出了我的预期。

I tried out with the following code, but the behavior is out of my expectation.

class Klass
   attr_accessor :array
end

s1 = Klass.new
ar = [1, 2, 3]
s1.array = [ar]

s2 = s1.clone
# according to the doc,
# s2.array should be initialized with empty Array
# however the array is recursivley copied too

s2.array.equal? s1.array # true

答

在Ruby中，所有对象都是参考。看下面的示例：

In Ruby, all objects are references. Take a look at the following example:

class Klass
  attr_accessor :a
end

s1 = Klass.new
a = [1,2,3]
s1.a = a
s2 = s1.clone
s1.a.object_id  #=> 7344240 
s2.a.object_id  #=> 7344240

您可以看到两个数组都是同一个对象，并且都是引用到位于堆中某处的数组。在深层副本中，该数组本身将被复制，而新的 s2 将具有其自己的独特数组。

You can see that both of the arrays are the same object, and are both references to the array living somewhere in the heap. In a deep copy, the array itself would have been copied, and the new s2 would have its own, distinct array. The array is not copied, just referenced.

注意：
如果进行深层复制，它的外观如下：

Note: Here's what it looks like if you do a deep copy:

s3 = Marshal.load(Marshal.dump(s1)) #=> #<Klass:0x00000000bf1350 @a=[1, 2, 3, 4], @bork=4> 
s3.a << 5 #=> [1, 2, 3, 4, 5] 
s1 #=> #<Klass:0x00000000e21418 @a=[1, 2, 3, 4], @bork=4>

如何理解#dup和#clone对引用其他对象的对象进行操作？

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