Question

我正在尝试读取和编写Guids数组到AppFabric缓存。我的探查器向我显示它将此序列化为Xml，这意味着事情进展得太慢了。 Guid[20000]需要60 ms才能添加到缓存，而类似大小的int[80000]需要10 ms。我想要一个Guids数组肯定在内部看起来像一个字节数组。达到这个目标的最快方法是什么，同时产生尽可能少的meta-fluff？我知道我将在缓存中添加和访问，并且数据不会特别持久，因此我不关心序列化类信息。

Answer 1

如果AppFabric缓存使用的序列化使用了除WCF的二进制编写器以外的任何东西，我会感到惊讶，所以如果是这种情况（看起来像这样），那么差异可以通过对数组的处理来解释二进制格式。某些基本类型的数组具有special node type，允许它们以二进制格式非常有效地存储。 int（和byte）是其中一些类型。 Guid不是（我不知道团队选择决定什么是“阵列类型”）。因此，虽然以二进制格式序列化整数数组非常有效，但序列化Guid数组并非如此。

正如m0sa建议的那样，你可以将它转换为byte []，你可能会看到一个很大的改进。我发现LINQ语法虽然更清晰，但并没有为您提供更加传统的for循环所带来的巨大性能提升。我运行下面的代码来比较序列化速度（我认为它将映射到AF缓存），并且序列化为byte []（即使在Guid []到byte []之间进行转换）也比来自int []的那个。

public class StackOverflow_6346646
{
    static void SerializeGuid()
    {
        Console.WriteLine("Serializing Guid[]");
        var guids = new Guid[20000];
        Random rndGen = new Random();
        for (int i = 0; i < guids.Length; i++)
        {
            guids[i] = Guid.NewGuid();
        }
        MemoryStream ms = new MemoryStream();
        Stopwatch watch = new Stopwatch();
        DataContractSerializer dcs = new DataContractSerializer(guids.GetType());
        XmlDictionaryWriter binaryWriter = XmlDictionaryWriter.CreateBinaryWriter(ms);
        watch.Start();
        dcs.WriteObject(binaryWriter, guids);
        binaryWriter.Flush();
        watch.Stop();
        Console.WriteLine("Serialized in {0}ms, total size = {1} bytes", watch.ElapsedMilliseconds, ms.Position);
    }
    static void SerializeInt()
    {
        Console.WriteLine("Serializing int[]");
        var guids = new int[80000]; // new Guid[20000];
        Random rndGen = new Random();
        for (int i = 0; i < guids.Length; i++)
        {
            guids[i] = rndGen.Next(); // Guid.NewGuid();
        }
        MemoryStream ms = new MemoryStream();
        Stopwatch watch = new Stopwatch();
        DataContractSerializer dcs = new DataContractSerializer(guids.GetType());
        XmlDictionaryWriter binaryWriter = XmlDictionaryWriter.CreateBinaryWriter(ms);
        watch.Start();
        dcs.WriteObject(binaryWriter, guids);
        binaryWriter.Flush();
        watch.Stop();
        Console.WriteLine("Serialized in {0}ms, total size = {1} bytes", watch.ElapsedMilliseconds, ms.Position);
    }
    static void SerializeGuidAsByteArray(bool useLinq)
    {
        Console.WriteLine("Serializing Guid[] as byte[], {0}", useLinq ? "using LINQ" : "not using LINQ");
        var guids = new Guid[20000];
        Random rndGen = new Random();
        for (int i = 0; i < guids.Length; i++)
        {
            guids[i] = Guid.NewGuid();
        }

        MemoryStream ms = new MemoryStream();
        Stopwatch watch = new Stopwatch();
        DataContractSerializer dcs = new DataContractSerializer(typeof(byte[]));
        XmlDictionaryWriter binaryWriter = XmlDictionaryWriter.CreateBinaryWriter(ms);
        watch.Start();
        byte[] bytes;
        if (useLinq)
        {
            bytes = guids.SelectMany(x => x.ToByteArray()).ToArray();
        }
        else
        {
            bytes = new byte[guids.Length * 16];
            for (int i = 0; i < guids.Length; i++)
            {
                byte[] guidBytes = guids[i].ToByteArray();
                Buffer.BlockCopy(guidBytes, 0, bytes, 16 * i, 16);
            }
        }
        dcs.WriteObject(binaryWriter, bytes);
        binaryWriter.Flush();
        watch.Stop();
        Console.WriteLine("Serialized in {0}ms, total size = {1} bytes", watch.ElapsedMilliseconds, ms.Position);
    }
    public static void Test()
    {
        SerializeGuid();
        SerializeInt();
        SerializeGuidAsByteArray(true);
        SerializeGuidAsByteArray(false);
    }
}

Answer 2

使用Guid。ToByteArray（）

Guid[] yourArray = ...;
byte[][] serializedArray = yourArray.Select(x => x.ToByteArray()).ToArray();

你甚至可以将它序列化为一维数组，因为Guid字节数组的长度是已知的（16）：

byte[] serializedArray = yourArray.SelectMany(x => x.ToByteArray()).ToArray();

将Guid []快速序列化为二进制文件以存储在C＃AppFabric中

2 个答案: