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推荐答案
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>更多推荐
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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