我在 Google 中随机输入“最快的序列化 c#”,结果得到了 protobuf.net。我试过了,我认为我可以正确序列化,但由于我无法反序列化,现在无法判断是否存在?!
当试图去序列化时,我得到:
A first chance exception of type 'ProtoBuf.ProtoException' occurred in protobuf-net.dll
凉爽的。
要序列化的数据:
[ProtoContract]
public struct Cow
{
[ProtoMember(1)]
public float Weight{ get; private set; }
[ProtoMember(2)]
public bool[] HadCowlings{ get; private set; }
public Cow(float weight, bool[] babies)
: this()
{
this.Weight = weight;
this.HadCowlings= (bool[])babies.Clone();
}
...
}
[ProtoContract]
public class Pasture
{
[ProtoMember(1)]
public Point Position { get; private set; }
[ProtoMember(2)]
public Cow[] Cows { get; private set; }
public static int HerdSize { get; private set; }
public static float BoundWidth { get; private set;}
public static float BoundHeight { get; private set; }
public Pasture(Cow[] Cows, Point farmPosition)
{
this.Cows = (Cow[])Cows.Clone();
Position = farmPosition;
}
...
}
[ProtoContract]
public class Farm
{
[ProtoMember(1)]
public Point FarmIDCoordinates{ get; private set; }
[ProtoMember(2)]
public List<Pasture> Pastures{ get; private set; }
public static float BoundWidth { get; private set; }
public static float BoundHeight { get; private set; }
public static int FarmSize { get; private set; }
public Farm(int x, int y, FarmType fType)
{
if (fType == RegionType.STANDARD)
Pastures = new List<Pasture>(//make a farm!);
else
Pastures = new List<Pasture>(//What he said);
FarmIDCoordinates = new Point(x, y);
}
...
}
如何:
放:
using (ObjectSerializer serializer = new ObjectSerializer())
{
serializer.ProtoSerialize<Farm>(farm.ToString() + ".bin", aFarm)
}
得到:
using (ObjectSerializer serializer = new ObjectSerializer())
{
try
{
farmsIOwn.Add(serializer.ProtoDeserialize<Farm>(
farmLat.X.ToString() + "_" + farmLong.Y.ToString() + ".bin"));
}
catch
{
// make me a dummy farm, crashing is for dummies
}
}
对象序列化器:
public void ProtoSerialize<T>(string fileName, T objectGraph)
{
using (var stream = File.Open(fileName, FileMode.Create))
{
Serializer.Serialize<T>(stream, objectGraph);
}
}
public T ProtoDeserialize<T>(string fileName)
{
T objectGraph;
using (var stream = File.Open(fileName, FileMode.Open))
{
objectGraph = Serializer.Deserialize<T>(stream);
}
return objectGraph;
}