When you deserialize an object, the transport format determines whether you will create a stream or file object. After the transport format is determined, you can call the Serialize or Deserialize methods, as required.
To deserialize an object
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Construct a XmlSerializer using the type of the object to deserialize.
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Call the Deserialize method to produce a replica of the object. When deserializing, you must cast the returned object to the type of the original, as shown in the following example, which deserializes the object into a file (although it could also be deserialized into a stream).
Dim myObject As MySerializableClass
' Construct an instance of the XmlSerializer with the type
' of object that is being deserialized.
Dim mySerializer As XmlSerializer = New XmlSerializer(GetType(MySerializableClass))
' To read the file, create a FileStream.
Dim myFileStream As FileStream = _
New FileStream("myFileName.xml", FileMode.Open)
' Call the Deserialize method and cast to the object type.
myObject = CType( _
mySerializer.Deserialize(myFileStream), MySerializableClass)
MySerializableClass myObject;
// Construct an instance of the XmlSerializer with the type
// of object that is being deserialized.
XmlSerializer mySerializer =
new XmlSerializer(typeof(MySerializableClass));
// To read the file, create a FileStream.
FileStream myFileStream =
new FileStream("myFileName.xml", FileMode.Open);
// Call the Deserialize method and cast to the object type.
myObject = (MySerializableClass)
mySerializer.Deserialize(myFileStream)
See Also