c# - 有没有办法获取 System.Random 的实际状态？

Question

我希望能够获得 System.Random 的实际状态或种子或其他任何东西，这样我就可以关闭一个应用程序，当用户重新启动它时，它只是用存储的那个“重新播种”它并像以前一样继续从未关闭。

这可能吗？

根据 Jon 的想法，我想出了这个来测试它；

static void Main(string[] args)
{
    var obj = new Random();
    IFormatter formatter = new BinaryFormatter();
    Stream stream = new FileStream("c:\\test.txt", FileMode.Create, FileAccess.Write, FileShare.None);
    formatter.Serialize(stream, obj);
    stream.Close();
    for (var i = 0; i < 10; i++)
        Console.WriteLine(obj.Next().ToString());

    Console.WriteLine();

    formatter = new BinaryFormatter();
    stream = new FileStream("c:\\test.txt", FileMode.Open, FileAccess.Read, FileShare.Read);
    obj = (Random)formatter.Deserialize(stream);
    stream.Close();
    for (var i = 0; i < 10; i++)
        Console.WriteLine(obj.Next().ToString());

    Console.Read();
}

Answer 1

它是可序列化的，因此您可能会发现只需使用 BinaryFormatter 并保存字节数组...

示例代码:

using System;
using System.IO;
using System.Runtime.Serialization.Formatters.Binary;

public class Program
{
    public static void Main(String[] args)
    {
        Random rng = new Random();
        Console.WriteLine("Values before saving...");
        ShowValues(rng);

        BinaryFormatter formatter = new BinaryFormatter(); 
        MemoryStream stream = new MemoryStream();
        formatter.Serialize(stream, rng);

        Console.WriteLine("Values after saving...");
        ShowValues(rng);

        stream.Position = 0; // Rewind ready for reading
        Random restored = (Random) formatter.Deserialize(stream);

        Console.WriteLine("Values after restoring...");
        ShowValues(restored);       
    }

    static void ShowValues(Random rng)
    {
        for (int i = 0; i < 5; i++)
        {
            Console.WriteLine(rng.Next(100));
        }
    }
}

sample 运行的结果很有希望:

Values before saving...
25
73
58
6
33
Values after saving...
71
7
87
3
77
Values after restoring...
71
7
87
3
77

不可否认，我并不热衷于内置序列化，但如果这是一个相当快速和肮脏的 hack，那应该没问题......