I am trying out the AsReference option in ProtoMember for recursive referencing. If I create AnOwner with the public constructor and then serialize/deserialize, AnOwner.Data becomes null. Can someone explain what happens internally and whether recursive referencing is supported? Thanks!
[ProtoContract()]
public class SomeData
{
[ProtoMember(1, AsReference = true)]
public AnOwner Owner;
[ProtoMember(2)]
public string Value;
/// <summary>
/// ProtoBuf deserialization constructor. In fact, Serializer did not complain when this is missing
/// </summary>
private SomeData()
{
}
public SomeData(string value)
{
Value = value;
}
}
[ProtoContract()]
public class AnOwner
{
[ProtoMember(1)]
public SomeData Data;
/// <summary>
/// ProtoBuf deserialization constructor
/// </summary>
private AnOwner()
{
}
public AnOwner(SomeData data)
{
Data = data;
Data.Owner = this;
}
}
EDIT: After much deliberation, I manage to understand it in the form of this small demo which I will share here. With the current implementation (v2 beta) it matters if AsReference=true is specified for both, neither or either with respect to which object is passed into Serializer.Serialize().
public class Program
{
using System.IO;
using ProtoBuf;
using System;
public static void main();
{
AnOwner owner1, owner2;
AnOwner owner = new AnOwner();
SomeData data = new SomeData();
owner.Data = data;
data.Owner = owner;
string file = "sandbox.txt";
try { File.Delete(file); } catch {}; // Just in case, cos' it felt like some caching was in place.
using (var fs = File.OpenWrite(file)) { Serializer.Serialize(fs, owner); }
using (var fs = File.OpenRead(file)) { owner1 = Serializer.Deserialize<AnOwner>(fs); }
using (var fs = File.OpenRead(file)) { owner2 = Serializer.Deserialize<AnOwner>(fs); }
Console.WriteLine("SomeData.i: {0}, {1}, {2}, {3}", owner1.Data.i, owner1.Data.Owner.Data.i, owner2.Data.i, owner2.Data.Owner.Data.i);
Console.WriteLine("AnOwner.i: {0}, {1}, {2}, {3}", owner1.i, owner1.Data.Owner.i, owner2.i, owner2.Data.Owner.i);
System.Diagnostics.Debug.Assert(owner1 == owner1.Data.Owner, "1. Expect reference same, but not the case.");
System.Diagnostics.Debug.Assert(owner2 == owner2.Data.Owner, "2. Expect reference same, but not the case.");
System.Diagnostics.Debug.As开发者_如何学Csert(owner1 != owner2, "3. Expect reference different, but not the case.");
}
}
[ProtoContract()]
public class SomeData
{
public static readonly Random RAND = new Random(2);
[ProtoMember(1, AsReference = true)]
public AnOwner Owner;
// Prove that SomeData is only instantiated once per deserialise
public int i = RAND.Next(100);
public SomeData() { }
}
[ProtoContract()]
public class AnOwner
{
public static readonly Random RAND = new Random(3);
[ProtoMember(1, AsReference=true)]
public SomeData Data;
// Prove that AnOwner is only instantiated once per deserialise
public int i = RAND.Next(100);
/// <summary>
/// ProtoBuf deserialization constructor
/// </summary>
public AnOwner() { }
}
Basically, instead of serializing AnOwner directly, it serializes a fake (doesn't really exist) object with one or more of the following:
- the existing key (integer) to an object that has already been seen
- a new key
- the object itself
- the type information (if DynamicType is enabled)
When serializing a tracked object, an internal list is checked; if the object is there (has been seen before), then the old key (only) is written. Otherwise a new key is generated and stored against that object, and the new key and the object are written. When deserializing, if a "new key" is found, the object data is deserialized and the new object is stored against that key (actually the order here is a bit complex, to handle recursion). If an "old key" is found, the internal list is used to fetch the old old object.
For almost all objects, comparison is on a reference equality basis (even if equality is overriden). Note that this works slightly differently for strings, which are compared for string equality - so two different instances of the string "Fred"
will still share a key.
I believe most recursion scenarios are supported, but if you get an issue please let me know.
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