Adding a clone of an object to a list in c# (prevent modification from outside)

Question

I have an object like 'obj1' which I want to add to a list. I can add it by just list1.add(obj1) .Now once I update obj1, the object in my list is also updating! (I understand that I am dealing with references here)

My requirement demands modifying the obj1 and add it to the list again! Instead of having two different objects, I have only one, because for both of them, the reference is the same - obj1.

Is there any way I can modify this obj1 and add it to the list and still not lose the old one? Any workarounds would be extremely helpful!

Thanks in Advance!

Answer 1

The C# language does not support cloning of objects. Therefore, if obj1 is not a value object (i.e. a struct), you cannot do that. Note: there is the possibility of implementing ICloneable, however, its use is not advised.

One approach I use in another project is to use AutoMapper to create a copy of the object before inserting into the list. Example:

MyType copy = Mapper.DynamicMap(obj1);
list.Add(copy);

Please use that approach for value holder types only, especially not for types that implement IDisposable or something similar.

Answer 2

我找到了一种使用AutoMapper的方法。虽然AutoMapper看起来像一个很棒的工具，但是我无法在需要Net2.0 / Mono的项目中使用它。

您可以使用序列化程序来创建对象的副本/克隆。我使用json.NET是因为我已经在项目中使用它了，但是我想其他库也可以使用。基本上，您将对象序列化为一个字符串，然后从该字符串创建一个新对象，因为该字符串未绑定到原始对象，所以您将获得一个全新的对象。

这是使用json.net的代码示例：

List<SomeObject> list1 = new List<SomeObject>();
SomeObject obj1 = new SomeObject(params, etc);

string data = JsonConvert.SerializeObject(obj1);
SomeObject obj2 = JsonConvert.DeserializeObject<SomeObject>(data);
list1.Add(obj2);

您甚至可以将最后三行缩短为以下形式：

list1.Add(JsonConvert.DeserializeObject<SomeObject>(JsonConvert.SerializeObject(obj1)));

您可能可以编写一个函数/方法来为您执行此操作。由于我创建自己的对象类型，因此向该对象添加了一个对象，如下所示：

public ItemInputData copyOf()
{
    string data = JsonConvert.SerializeObject(this);
    ItemInputData copy = JsonConvert.DeserializeObject<ItemInputData>(data);
    return copy;
}

用于将副本添加到列表的代码如下所示：

list1.Add(item.copyOf());

希望这会帮助=]