为什么 String::sub!() 会改变 Ruby 中克隆对象的原始对象？
我的 Ruby 代码中有一个结构，稍后看起来有点像这样
Parameter = Struct.new(:name, :id, :default_value, :minimum, :maximum)
，我使用创建此结构的实例
freq = Parameter.new('frequency', 15, 1000.0, 20.0, 20000.0)
在某些时候，我需要此结构的精确副本，所以我调用
newFreq = freq.clone
然后，我更改 newFreq > 的名字
newFreq.name.sub!('f', 'newF')
也奇迹般地改变了 freq.name！
像 newFreq.name = 'newFrequency' 这样的简单赋值不会改变 freq。
这是应该的工作方式吗？
编辑：使用类而不是结构并重载clone来进行深层复制是个好主意吗？
I have a struct in my Ruby code that looks somewhat like this
Parameter = Struct.new(:name, :id, :default_value, :minimum, :maximum)
later, I create an instance of this struct using
freq = Parameter.new('frequency', 15, 1000.0, 20.0, 20000.0)
At some point, I need an exact duplicate of this struct, so I call
newFreq = freq.clone
Then, I change newFreq's name
newFreq.name.sub!('f', 'newF')
Which, miraculously, changes freq.name, too!
A simple assignment like newFreq.name = 'newFrequency' does not change freq.
Is this the way this is supposed to work?
Edit: Is it a good idea to use a class instead of a struct and overload clone to make a deep copy?
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newFreq是freq的浅拷贝。这意味着newFreq中存储的每个引用都指向存储在freq中的对象。您可以独立更改引用指向的位置 (newFreq.name = newFreq.name.sub 'f','newF')，但如果您调用改变对象的方法，则newFreq和freq将受到影响。另请参阅http://en.wikipedia.org/wiki/Object_copy
newFreqis a shallow copy offreq. That means that each of the references stored inside ofnewFreqpoints to the object as the ones stored infreq. You can change where the references point independantly (newFreq.name = newFreq.name.sub 'f','newF'), but if you call a method that mutates the object, bothnewFreqandfreqwill be affected.See also http://en.wikipedia.org/wiki/Object_copy
Object#clone方法执行浅复制。您需要深复制才能完成工作。请点击此链接了解 Ruby 中的深度复制。
The
Object#clonemethod performs shallow copy. You need deep copy to get the job done.Follow this link to learn about deep copy in Ruby.