blocks|key|1541204|text|实际上，List是对B元素的引用列表。引用列表已更改，但引用仍指向相同的对象。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1541205|如果您需要克隆对象，请查看此topic。|1541206|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/5359318/how-to-clone-objects^0|4|7|0|E|5|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@$9|T|A|U|1|V]]|E|$]]|$1|H|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

Actually <code>List&lt;B&gt;</code> is a list of reference to B elements. The List of reference changed, but the references still point to the same objects. 

If a clone object is what you need, take a look at this <a href="https://stackoverflow.com/questions/5359318/how-to-clone-objects">topic</a>.

blocks|key|3856629|text|除了简单类型(整型、双精度等)，C#中的所有类型都作为引用传递。如果要创建副本，则需要手动执行此操作|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3856630|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

All types in C# are passed as reference except simple types (int, double etc.). If you want to create a copy you need to do this manually

blocks|key|5324650|text|当b.Where(..)时，.ToList则only+copy+is+created.Where返回IEnumerable，这允许您使用foreach循环进行迭代。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5324651|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

When b.Where(..).ToList then only copy is created.Where returns IEnumerable which allows you to iterate using foreach loop.

blocks|key|5324700|text|Where(t+=>+IsOk(t));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5324701|不会创建调用它的IEnumerable的深层副本。它遍历集合并返回符合isOk()条件的项。这意味着它创建了一个新的列表，但条目是相同的。|unstyled|offset|length|style|CODE|5324702|如果您想要创建每个项目的副本，则应在这些项目上实现ICloneable，并将其克隆到新列表中。|5324703|entityMap^0|0|8|B|0|P|A|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@$G|T|H|U|I|J]]|9|@]|A|$]]|$1|M|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>Where(t =&gt; IsOk(t));
</code></pre>

doesn't create a deep copy of the <code>IEnumerable</code> on which it's called. It iterates over the collection and returns the items matching the isOk() criteria. Meaning it creates a new list but the items are the same.

If you would like to create a copy of the each of the items you should implement <code>ICloneable</code> on the items and Clone them in a new list.

blocks|key|1366722|text|你实际上是在做浅层复制:你复制对对象的引用，而不是创建对象的克隆：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1366723|++//+Shallow+copy+(your+actual+implementation):
++//+NewList+is+not+b,+but+it+contains+references+to+objects+in+b
++var+NewList+=+b
++++.Where(item+=>+...)
++++.ToList();+

++//+Deep+copy+(which+probably+you're+looking+for)
++//+NewList+doesn't+contain+any+reference+to+any+objects+in+b+
++var+NewList+=+b
++++.Where(item+=>+...)
++++.Select(item+=>+item.Clone())+//+You+need+this,+proving+that+Clone()+makes+a+deep+copy
++++.ToList();+|code-block|syntax|javascript|1366724|如果你想克隆这些项目，你必须为B类正确地实现ICloneable接口(以确保深度复制)|offset|length|style|CODE|1366725|entityMap^0|0|0|F|1|M|A|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

You;re actually doing shallow coping: you copy references to objects, without creating clones of objects:

<pre><code> // Shallow copy (your actual implementation):
 // NewList is not b, but it contains references to objects in b
 var NewList = b
 .Where(item =&gt; ...)
 .ToList(); 

 // Deep copy (which probably you're looking for)
 // NewList doesn't contain any reference to any objects in b 
 var NewList = b
 .Where(item =&gt; ...)
 .Select(item =&gt; item.Clone()) // You need this, proving that Clone() makes a deep copy
 .ToList(); 
</code></pre>

If you want to clone the items, you have to implement properly <code>ICloneable</code> interface for <code>B</code> class (to ensure the deep coping)

blocks|key|1366742|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1366743|为什么我的列表b改变了它的值？|blockquote|1366744|1366745|1366746|因为尽管您分配了一个新列表，但它们现在仍然将指向相同的+B+实例。在将新实例添加到新列表之前，您尚未创建新实例。|offset|length|style|BOLD|CODE|1366747|如何克隆现有对象的一个示例是通过复制构造函数|1366748|public+class+B
{
++++public+string+Name+{+get;+}
++++public+B(B+other)
++++{
++++++++Name+=+other.name;
++++}
}|code-block|syntax|javascript|1366749|现在，当您想要创建列表时，您需要执行以下操作：|1366750|foreach(var+elementOfB+in+b.Where(SomePredicate))
{
++++elementOfB.Name+=+"...";
++++newList.Add(new+B(elementOfB));
}|1366751|entityMap^0|0|0|0|0|M|5|S|1|U|2|S|1|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|Z|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|10|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|11|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|12|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|13|7|@$H|14|I|15|J|K]|$H|16|I|17|J|K]|$H|18|I|19|J|K]|$H|1A|I|1B|J|L]]|8|@]|9|$]]|$1|M|3|N|4|5|6|1C|7|@]|8|@]|9|$]]|$1|O|3|P|4|Q|6|1D|7|@]|8|@]|9|$R|S]]|$1|T|3|U|4|5|6|1E|7|@]|8|@]|9|$]]|$1|V|3|W|4|Q|6|1F|7|@]|8|@]|9|$R|S]]|$1|X|3|-4|4|5|6|1G|7|@]|8|@]|9|$]]]|Y|$]]

<blockquote>
 why my list b changed it's values?
</blockquote>

Because although you allocated a new list, both of them now still point to the same <code>B</code> instance. You haven't created a new instance prior to adding them to the new list.

An example of how you could clone an existing object, is via a copy constructor:

<pre><code>public class B
{
 public string Name { get; }
 public B(B other)
 {
 Name = other.name;
 }
}
</code></pre>

Now when you want to create the list, you do:

<pre><code>foreach(var elementOfB in b.Where(SomePredicate))
{
 elementOfB.Name = "...";
 newList.Add(new B(elementOfB));
}
</code></pre>

I have 2 object's let's call them <code>A</code>, <code>B</code> and method 

<pre><code>List&lt;B&gt; DoSomething(ref A a, List&lt;B&gt; b)
{
 List&lt;B&gt; newList = new List&lt;B&gt;();

 //
 //Doing something to ref A
 //

 foreach(var elementOfB in b.where(...))
 {
 //
 elementOfB.Name = "...";
 //
 newList.Add(elementOfB);
 }
 return newList;
}
</code></pre>

So, after that method is done my original list <code>b</code> have changed value <code>Name</code> field (<code>string</code>) but I did not pass it as <code>ref</code> and I'm using <code>where</code> that should return copy of elemens right? So my question is why my list <code>b</code> changed it's values?

<pre><code>List&lt;B&gt; originalList = ...;
List&lt;B&gt; newList = DoSomething(ref a, originalList);
//now originalList have changed Name field values
</code></pre>

Generic list and value reference

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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