我想做一些事情,比如:
MyObject myObj = GetMyObj(); // Create and fill a new object
MyObject newObj = myObj.Clone();
然后对未反映在原始对象中的新对象进行更改。
我通常不需要这个功能,所以当有必要时,我会创建一个新对象,然后单独复制每个属性,但这总是让我觉得有更好或更优雅的方法来处理这种情况。
如何克隆或深度复制对象,以便在原始对象中不反映任何更改的情况下修改克隆的对象?
我想做一些事情,比如:
MyObject myObj = GetMyObj(); // Create and fill a new object
MyObject newObj = myObj.Clone();
然后对未反映在原始对象中的新对象进行更改。
我通常不需要这个功能,所以当有必要时,我会创建一个新对象,然后单独复制每个属性,但这总是让我觉得有更好或更优雅的方法来处理这种情况。
如何克隆或深度复制对象,以便在原始对象中不反映任何更改的情况下修改克隆的对象?
当前回答
我已经创建了一个可接受答案的版本,该版本同时适用于“[Serializable]”和“[DataContract]”。我写它已经有一段时间了,但如果我没有记错,[DataContract]需要一个不同的序列化程序。
需要System、System.IO、System.Runtime.Serialization、System.Runtime.Serialization.Formatters.Binary、System.Xml;
public static class ObjectCopier
{
/// <summary>
/// Perform a deep Copy of an object that is marked with '[Serializable]' or '[DataContract]'
/// </summary>
/// <typeparam name="T">The type of object being copied.</typeparam>
/// <param name="source">The object instance to copy.</param>
/// <returns>The copied object.</returns>
public static T Clone<T>(T source)
{
if (typeof(T).IsSerializable == true)
{
return CloneUsingSerializable<T>(source);
}
if (IsDataContract(typeof(T)) == true)
{
return CloneUsingDataContracts<T>(source);
}
throw new ArgumentException("The type must be Serializable or use DataContracts.", "source");
}
/// <summary>
/// Perform a deep Copy of an object that is marked with '[Serializable]'
/// </summary>
/// <remarks>
/// Found on http://stackoverflow.com/questions/78536/cloning-objects-in-c-sharp
/// Uses code found on CodeProject, which allows free use in third party apps
/// - http://www.codeproject.com/KB/tips/SerializedObjectCloner.aspx
/// </remarks>
/// <typeparam name="T">The type of object being copied.</typeparam>
/// <param name="source">The object instance to copy.</param>
/// <returns>The copied object.</returns>
public static T CloneUsingSerializable<T>(T source)
{
if (!typeof(T).IsSerializable)
{
throw new ArgumentException("The type must be serializable.", "source");
}
// Don't serialize a null object, simply return the default for that object
if (Object.ReferenceEquals(source, null))
{
return default(T);
}
IFormatter formatter = new BinaryFormatter();
Stream stream = new MemoryStream();
using (stream)
{
formatter.Serialize(stream, source);
stream.Seek(0, SeekOrigin.Begin);
return (T)formatter.Deserialize(stream);
}
}
/// <summary>
/// Perform a deep Copy of an object that is marked with '[DataContract]'
/// </summary>
/// <typeparam name="T">The type of object being copied.</typeparam>
/// <param name="source">The object instance to copy.</param>
/// <returns>The copied object.</returns>
public static T CloneUsingDataContracts<T>(T source)
{
if (IsDataContract(typeof(T)) == false)
{
throw new ArgumentException("The type must be a data contract.", "source");
}
// ** Don't serialize a null object, simply return the default for that object
if (Object.ReferenceEquals(source, null))
{
return default(T);
}
DataContractSerializer dcs = new DataContractSerializer(typeof(T));
using(Stream stream = new MemoryStream())
{
using (XmlDictionaryWriter writer = XmlDictionaryWriter.CreateBinaryWriter(stream))
{
dcs.WriteObject(writer, source);
writer.Flush();
stream.Seek(0, SeekOrigin.Begin);
using (XmlDictionaryReader reader = XmlDictionaryReader.CreateBinaryReader(stream, XmlDictionaryReaderQuotas.Max))
{
return (T)dcs.ReadObject(reader);
}
}
}
}
/// <summary>
/// Helper function to check if a class is a [DataContract]
/// </summary>
/// <param name="type">The type of the object to check.</param>
/// <returns>Boolean flag indicating if the class is a DataContract (true) or not (false) </returns>
public static bool IsDataContract(Type type)
{
object[] attributes = type.GetCustomAttributes(typeof(DataContractAttribute), false);
return attributes.Length == 1;
}
}
其他回答
当使用Marc Gravells protobuf net作为序列化程序时,接受的答案需要一些轻微的修改,因为要复制的对象不会被归因于[Serializable],因此不可序列化,Clone方法将抛出异常。我修改了它以与protobuf net一起使用:
public static T Clone<T>(this T source)
{
if(Attribute.GetCustomAttribute(typeof(T), typeof(ProtoBuf.ProtoContractAttribute))
== null)
{
throw new ArgumentException("Type has no ProtoContract!", "source");
}
if(Object.ReferenceEquals(source, null))
{
return default(T);
}
IFormatter formatter = ProtoBuf.Serializer.CreateFormatter<T>();
using (Stream stream = new MemoryStream())
{
formatter.Serialize(stream, source);
stream.Seek(0, SeekOrigin.Begin);
return (T)formatter.Deserialize(stream);
}
}
这将检查[ProtoControl]属性的存在,并使用protobufs自己的格式化程序来序列化对象。
这将把一个对象的所有可读写财产复制到另一个对象。
public class PropertyCopy<TSource, TTarget>
where TSource: class, new()
where TTarget: class, new()
{
public static TTarget Copy(TSource src, TTarget trg, params string[] properties)
{
if (src==null) return trg;
if (trg == null) trg = new TTarget();
var fulllist = src.GetType().GetProperties().Where(c => c.CanWrite && c.CanRead).ToList();
if (properties != null && properties.Count() > 0)
fulllist = fulllist.Where(c => properties.Contains(c.Name)).ToList();
if (fulllist == null || fulllist.Count() == 0) return trg;
fulllist.ForEach(c =>
{
c.SetValue(trg, c.GetValue(src));
});
return trg;
}
}
这是你使用它的方式:
var cloned = Utils.PropertyCopy<TKTicket, TKTicket>.Copy(_tmp, dbsave,
"Creation",
"Description",
"IdTicketStatus",
"IdUserCreated",
"IdUserInCharge",
"IdUserRequested",
"IsUniqueTicketGenerated",
"LastEdit",
"Subject",
"UniqeTicketRequestId",
"Visibility");
或复制所有内容:
var cloned = Utils.PropertyCopy<TKTicket, TKTicket>.Copy(_tmp, dbsave);
如果您的对象树是可序列化的,您也可以使用以下内容
static public MyClass Clone(MyClass myClass)
{
MyClass clone;
XmlSerializer ser = new XmlSerializer(typeof(MyClass), _xmlAttributeOverrides);
using (var ms = new MemoryStream())
{
ser.Serialize(ms, myClass);
ms.Position = 0;
clone = (MyClass)ser.Deserialize(ms);
}
return clone;
}
请注意,此解决方案非常简单,但它不像其他解决方案那样高效。
请确保,如果类增长,仍然只有那些字段被克隆,这些字段也会被序列化。
保持简单,并使用AutoMapper,正如其他人提到的,这是一个简单的小库,可以将一个对象映射到另一个对象。。。要将一个对象复制到另一个具有相同类型的对象,只需要三行代码:
MyType source = new MyType();
Mapper.CreateMap<MyType, MyType>();
MyType target = Mapper.Map<MyType, MyType>(source);
目标对象现在是源对象的副本。不够简单?创建扩展方法以在解决方案中的任何地方使用:
public static T Copy<T>(this T source)
{
T copy = default(T);
Mapper.CreateMap<T, T>();
copy = Mapper.Map<T, T>(source);
return copy;
}
扩展方法可以如下使用:
MyType copy = source.Copy();
除了这里的一些精彩答案外,您可以在C#9.0及更高版本中完成以下操作(假设您可以将类转换为记录):
record Record
{
public int Property1 { get; set; }
public string Property2 { get; set; }
}
然后,只需使用with运算符将一个对象的值复制到新对象。
var object1 = new Record()
{
Property1 = 1,
Property2 = "2"
};
var object2 = object1 with { };
// object2 now has Property1 = 1 & Property2 = "2"
我希望这有助于:)