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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
Construct a XmlSerializer using the type of the object to deserialize.
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)