FileInfo fi = new FileInfo(xmlPATH);
Stream s =fi.Open(FileMode.Open, FileAccess.Read, FileShare.Read);
BinFormatter.Deserialize(s);
在BinFormatter.Deserialize(s);这一句出现异常
Message "输入流是无效的二进制格式。开始内容(以字节为单位)是: 73-22-3F-3E-0D-0A-3C-50-61-74-72-6F-6C-4D-6F-64-65..."
StackTrace " 在 System.Runtime.Serialization.Formatters.Binary.SerializationHeaderRecord.Read(__BinaryParser input)\r\n 在 System.Runtime.Serialization.Formatters.Binary.__BinaryParser.ReadSerializationHeaderRecord()\r\n 在 System.Runtime.Serialization.Formatters.Binary.__BinaryParser.Run()\r\n 在 System.Runtime.Serialization.Formatters.Binary.ObjectReader.Deserialize(HeaderHandler handler, __BinaryParser serParser, Boolean fCheck, Boolean isCrossAppDomain, IMethodCallMessage methodCallMessage)\r\n 在 System.Runtime.Serialization.Formatters.Binary.BinaryFormatter.Deserialize(Stream serializationStream, HeaderHandler handler, Boolean fCheck, Boolean isCrossAppDomain, IMethodCallMessage methodCallMessage)\r\n 在 System.Runtime.Serialization.Formatters.Binary.BinaryFormatter.Deserialize(Stream serializationStream)"