blocks|key|141141|text|您的目标变量是引用类型。这意味着您对它所做的任何操作都将反映在您传递给它的列表中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|141142|要不这样做，您需要在方法中创建一个新列表，将target内容复制到其中，然后在新列表上执行remove+at操作。|offset|length|style|CODE|141143|About+Reference+and+Value+Types|141144|entityMap|0|LINK|mutability|MUTABLE|url|http://cplus.about.com/od/learnc/ss/value.htm^0|0|M|6|0|0|V|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|V|8|@]|9|@$D|W|E|X|1|Y]]|A|$]]|$1|J|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

Your target variable is a reference type. This means that anything you do to it will be reflected in the list you pass into it. 

To not do that, you are going to need to create a new list in the method, copy <code>target</code> contents to it, and then perform the remove at operation on the new list. 

<a href="http://cplus.about.com/od/learnc/ss/value.htm" rel="nofollow">About Reference and Value Types</a>

blocks|key|4585307|text|您需要在您的方法中克隆列表，因为List<T>是一个类，所以它是引用类型的，并通过引用传递。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4585308|例如：|4585309|List<Item>+SomeOperationFunction(List<Item>+target)
{
++List<Item>+tmp+=+target.ToList();
++tmp.RemoveAt(3);
++return+tmp;
}|code-block|syntax|javascript|4585310|或|4585311|List<Item>+SomeOperationFunction(List<Item>+target)
{
++List<Item>+tmp+=+new+List<Item>(target);
++tmp.RemoveAt(3);
++return+tmp;
}|4585312|4585313|List<Item>+SomeOperationFunction(List<Item>+target)
{
++List<Item>+tmp+=+new+List<Item>();
++tmp.AddRange(target);
++tmp.RemoveAt(3);
++return+tmp;
}|4585314|entityMap^0|G|7|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Z|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|10|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|11|8|@]|D|@]|E|$K|L]]|$1|Q|3|N|5|6|7|12|8|@]|D|@]|E|$]]|$1|R|3|S|5|J|7|13|8|@]|D|@]|E|$K|L]]|$1|T|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|U|$]]

You need to clone your list in your method, because <code>List&lt;T&gt;</code> is a class, so it's reference-type and is passed by reference.

For example:

<pre><code>List&lt;Item&gt; SomeOperationFunction(List&lt;Item&gt; target)
{
 List&lt;Item&gt; tmp = target.ToList();
 tmp.RemoveAt(3);
 return tmp;
}
</code></pre>

Or

<pre><code>List&lt;Item&gt; SomeOperationFunction(List&lt;Item&gt; target)
{
 List&lt;Item&gt; tmp = new List&lt;Item&gt;(target);
 tmp.RemoveAt(3);
 return tmp;
}
</code></pre>

or

<pre><code>List&lt;Item&gt; SomeOperationFunction(List&lt;Item&gt; target)
{
 List&lt;Item&gt; tmp = new List&lt;Item&gt;();
 tmp.AddRange(target);
 tmp.RemoveAt(3);
 return tmp;
}
</code></pre>

blocks|key|4585331|text|首先构建一个新的list并对其进行操作，因为List是一个引用类型，即当您在函数中传递它时，您不仅传递了值，还传递了实际的对象本身。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4585332|如果只是将target赋值给mainList，则两个变量都指向同一个对象，因此您需要创建一个新列表：|offset|length|style|CODE|4585333|List<Item>+target+=+new+List<Item>(mainList);|code-block|syntax|javascript|4585334|void+List<Item>+SomeOperationFunction()没有任何意义，因为要么什么都不返回(void)，要么返回List<T>。因此，要么从您的方法中删除返回语句，要么返回一个新的List<Item>。在后一种情况下，我会将其重写为：|4585335|List<Item>+target+=+SomeOperationFunction(mainList);

List<Item>+SomeOperationFunction(List<Item>+target)
{
++++var+newList+=+new+List<Item>(target);
++++newList.RemoveAt(3);
++++return+newList;
}|4585336|entityMap^0|0|5|6|E|8|0|0|0|13|1L|4|1V|7|2T|A|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$D|10|E|11|F|G]|$D|12|E|13|F|G]|$D|14|E|15|F|G]|$D|16|E|17|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|18|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|R|$]]

Build a new list first and operate on that, because List is a reference type, i.e. when you pass it in a function, you do not just pass the value but the actual object itself.

If you just assign <code>target</code> to <code>mainList</code>, both variables point to the same object, so you need to create a new List:

<pre><code>List&lt;Item&gt; target = new List&lt;Item&gt;(mainList);
</code></pre>

<code>void List&lt;Item&gt; SomeOperationFunction()</code> makes no sense, because either you return nothing (<code>void</code>) or you return a <code>List&lt;T&gt;</code>. So either remove the return statement from your method or return a new <code>List&lt;Item&gt;</code>. In the latter case, I would rewrite this as:

<pre><code>List&lt;Item&gt; target = SomeOperationFunction(mainList);

List&lt;Item&gt; SomeOperationFunction(List&lt;Item&gt; target)
{
 var newList = new List&lt;Item&gt;(target);
 newList.RemoveAt(3);
 return newList;
}
</code></pre>

blocks|key|4103669|text|您需要制作列表的副本，以便对副本所做的更改不会影响原始列表。最简单的方法是在System.Linq中使用ToList扩展方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4103670|var+newList+=+SomeOperationFunction(target.ToList());|code-block|syntax|javascript|4103671|entityMap^0|12|B|1G|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You need to make a copy of the list so that changes to the copy won't affect the original. The easiest way to do that is to use the <code>ToList</code> extension method in <code>System.Linq</code>.

<pre><code>var newList = SomeOperationFunction(target.ToList());
</code></pre>

blocks|key|141196|text|您看到的是原始列表正在被修改，因为在默认情况下，任何非原语对象都是通过引用传递的(它实际上是通过值传递的，值就是引用，但这是另一回事)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|141197|您需要做的是克隆对象。这个问题将帮助您使用一些代码来克隆C#中的列表：How+do+I+clone+a+generic+list+in+C#?|offset|length|141198|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/q/222598/346561^0|0|Z|10|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@$D|P|E|Q|1|R]]|A|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]]]

Your are seeing the original list being modified because, by default, any non-primitive objects, are passed by reference (It is actually pass by value, the value being the reference, but that is a different matter).

What you need to do is clone the object. This question will help you with some code to clone a List in C#: <a href="https://stackoverflow.com/q/222598/346561">How do I clone a generic list in C#?</a>

blocks|key|4103854|text|因为List是引用类型，所以传递给函数的是对原始列表的引用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4103855|有关如何在C#中传递参数的详细信息，请参阅此MSDN+article。|4103856|为了实现您想要的结果，您应该在SomeOperationFunction中创建列表的副本，并返回此列表。一个简单的例子：|4103857|void+List<Item>+SomeOperationFunction(List<Item>+target)
{
++var+newList+=+new+List<Item>(target);
++newList.RemoveAt(3);
++return+newList;+//+return+copy+of+list
}|code-block|syntax|javascript|4103858|正如Olivier+Jacot-Descombes在对另一个答案的评论中指出的那样，重要的是要记住|4103859|4103860|+...如果项是引用类型，则列表仍保留对相同项的引用。因此，对项目本身的更改仍然会影响两个列表中的项目。|blockquote|4103861|4103862|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/0f66670z%2528v=vs.110%2529.aspx^0|2|4|0|M|C|0|0|F|L|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|16|8|@]|D|@$9|17|A|18|1|19]]|E|$]]|$1|H|3|I|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|J|3|K|5|L|7|1D|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]|$1|R|3|S|5|T|7|1G|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]|$1|V|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|W|$X|$5|Y|Z|10|E|$11|12]]]]

Since a <code>List</code> is a reference type, what is passed to the function is a reference to the original list.

See this <a href="http://msdn.microsoft.com/en-us/library/0f66670z%28v=vs.110%29.aspx" rel="nofollow">MSDN article</a> for more information about how parameters are passed in C#.

In order to achieve what you want, you should create a copy of the list in <code>SomeOperationFunction</code> and return this instead. A simple example:

<pre><code>void List&lt;Item&gt; SomeOperationFunction(List&lt;Item&gt; target)
{
 var newList = new List&lt;Item&gt;(target);
 newList.RemoveAt(3);
 return newList; // return copy of list
}
</code></pre>

As pointed out by Olivier Jacot-Descombes in the comments to another answer, it is important to bear in mind that

<blockquote>
 [...] the list still holds references to the same items if
 the items are of a reference type. So changes to the items themselves
 will still affect the items in both lists.
</blockquote>

blocks|key|141255|text|我不会将mainList赋值给目标，而是这样做：target.AddRange(mainList);|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|141256|然后，您将拥有这些项的副本，而不是对列表的引用。|141257|entityMap^0|O|Q|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

Instead of assigning mainList to target, I would do: <code>target.AddRange(mainList);</code>

Then you will have a copy of the items instead of a reference to the list.

blocks|key|753784|text|只需确保使用通过复制源列表的元素创建的列表初始化新列表即可。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|753785|List<Item>+target+=+mainList;应为List<item>+target+=+new+List<Item>(mainList);|offset|length|style|CODE|753786|entityMap^0|0|0|T|V|19|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

Just make sure that you initialize the new list with a list created by copying the elements of the source list.

<code>List&lt;Item&gt; target = mainList;</code> Should be <code>List&lt;item&gt; target = new List&lt;Item&gt;(mainList);</code>

blocks|key|4104009|text|您需要复制列表，因为在您的原始代码中，您所做的只是传递一个引用，正如您正确地怀疑的那样(有人会将其称为指针)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4104010|您可以调用新列表上的构造函数，将原始列表作为参数传递：|4104011|List<Item>+SomeOperationFunction(List<Item>+target)
{
++++List<Item>+result+=+new+List<Item>(target);
++++result.removeat(3);
++++return+result;
}|code-block|syntax|javascript|4104012|或创建MemberWiseClone|offset|length|4104013|List<Item>+SomeOperationFunction(List<Item>+target)
{
++++List<Item>+result+=+target.MemberWiseClone();
++++result.removeat(3);
++++return+result;
}|4104014|另外，您没有在任何地方存储SomeOperationFunction的返回值，因此您可能需要修改这一部分(您将该方法声明为void，该方法不应该返回任何内容，但在它内部返回的是一个对象)。您应该这样调用该方法：|style|CODE|4104015|List<Item>+target+=+SomeOperationFunction(mainList);|4104016|注意：不会复制列表中的元素(仅复制其引用)，因此修改元素的内部状态将影响两个列表。|BOLD|4104017|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/system.object.memberwiseclone.aspx^0|0|0|0|3|F|0|0|0|D|L|1P|4|0|0|0|3|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|16|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|17|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|18|8|@]|9|@$K|19|L|1A|1|1B]]|A|$]]|$1|M|3|N|5|F|7|1C|8|@]|9|@]|A|$G|H]]|$1|O|3|P|5|6|7|1D|8|@$K|1E|L|1F|Q|R]|$K|1G|L|1H|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|F|7|1I|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1J|8|@$K|1K|L|1L|Q|W]]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]]|Y|$Z|$5|10|11|12|A|$13|14]]]]

You'll need to make a copy of the list since in your original code what you're doing is just passing around, as you correctly suspected, a reference (someone would call it a pointer). 

You could either call the constructor on the new list, passing the original list as parameter:

<pre><code>List&lt;Item&gt; SomeOperationFunction(List&lt;Item&gt; target)
{
 List&lt;Item&gt; result = new List&lt;Item&gt;(target);
 result.removeat(3);
 return result;
}
</code></pre>

Or create a <a href="http://msdn.microsoft.com/en-us/library/system.object.memberwiseclone.aspx" rel="nofollow">MemberWiseClone</a>:

<pre><code>List&lt;Item&gt; SomeOperationFunction(List&lt;Item&gt; target)
{
 List&lt;Item&gt; result = target.MemberWiseClone();
 result.removeat(3);
 return result;
}
</code></pre>

Also, you are not storing the return of <code>SomeOperationFunction</code> anywhere, so you might want to revise that part as well (you declared the method as <code>void</code>, which should not return anything, but inside it you're returning an object).
You should call the method this way:

<pre><code>List&lt;Item&gt; target = SomeOperationFunction(mainList);
</code></pre>

Note: the elements of the list will not be copied (only their reference is copied), so modifying the internal state of the elements will affect both lists.

blocks|key|753855|text|即使您创建了一个新列表，对新列表中的项的引用仍将指向旧列表中的项，因此如果我需要一个包含新引用的新列表，我喜欢使用此扩展方法……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|753856|public+static+IEnumerable<T>+Clone<T>(this+IEnumerable<T>+target)+where+T+:+ICloneable
{
++++If+(target.IsNull())
++++++++throw+new+ArgumentException();

++++List<T>+retVal+=+new+List<T>();

++++foreach+(T+currentItem+in+target)
++++++++retVal.Add((T)(currentItem.Clone()));

++++return+retVal.AsEnumerable();
}|code-block|syntax|javascript|753857|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Even if you create a new list, the references to the items in the new list will still point to the items in the old list, so I like to use this extension method if I need a new list with new references...

<pre><code>public static IEnumerable&lt;T&gt; Clone&lt;T&gt;(this IEnumerable&lt;T&gt; target) where T : ICloneable
{
 If (target.IsNull())
 throw new ArgumentException();

 List&lt;T&gt; retVal = new List&lt;T&gt;();

 foreach (T currentItem in target)
 retVal.Add((T)(currentItem.Clone()));

 return retVal.AsEnumerable();
}
</code></pre>

blocks|key|753884|text|我尝试了上面的许多答案。在我测试的所有列表中，对新列表的更新都会修改原始列表。这就是对我有效的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|753885|var+newList+=+JsonConvert.DeserializeObject<List<object>>(JsonConvert.SerializeObject(originalList));
return+newlist.RemoveAt(3);|code-block|syntax|javascript|753886|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I tried many of the answers above. On all the ones I tested, an update to the new list modifies the original. this is what works for me.

<pre><code>var newList = JsonConvert.DeserializeObject&lt;List&lt;object&gt;&gt;(JsonConvert.SerializeObject(originalList));
return newlist.RemoveAt(3);
</code></pre>

I have a list that gets filled in with some data from an operation and I am storing it in the memory cache. Now I want another list which contains some sub data from the list based on some condition. 

As can be seen in the below code I am doing some operation on the target list. The problem is that whatever changes I am doing to the target list is also being done to the mainList. I think its because of the reference is same or something.

All I need is that operation on the target list not affect data inside the main list.

<pre><code>List&lt;Item&gt; target = mainList;
SomeOperationFunction(target);

 void List&lt;Item&gt; SomeOperationFunction(List&lt;Item&gt; target)
{
 target.removeat(3);
 return target;
}
</code></pre>

How do I change my new list without changing the original list?

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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我有一个列表，其中填充了来自操作的一些数据，并将其存储在内存缓存中。现在我想要另一个列表，它包含列表中基于某些条件的一些子数据。从下面的代码中可以看出，我正在对目标列表进行一些操作。问题是，无论我对目标列表做了什么更改，都会对mainList进行更改。我认为这是因为引用是相同的或什么的。我所需要的就是目标列表上的操作不...

问如何在不更改原始列表的情况下更改新列表？
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问如何在不更改原始列表的情况下更改新列表？EN