在我的下面的代码中,我锁定guid,尝试使其线程安全。 使用我的示例应用程序,我将获得一个"重复键"大约每10次我运行该程序。阿卡,我得到了一份副本,这不是我需要的。
无论如何都要制作" .NextGuid"线程安全?
using System;
namespace MyConsoleOne.BAL
{
public class GuidStore
{
private static object objectlock = new object();
private Guid StartingGuid { get; set; }
private Guid? LastGuidHolder { get; set; }
public GuidStore(Guid startingGuid)
{
this.StartingGuid = startingGuid;
}
public Guid? GetNextGuid()
{
lock (objectlock)
{
if (this.LastGuidHolder.HasValue)
{
this.LastGuidHolder = Increment(this.LastGuidHolder.Value);
}
else
{
this.LastGuidHolder = Increment(this.StartingGuid);
}
}
return this.LastGuidHolder;
}
private Guid Increment(Guid guid)
{
byte[] bytes = guid.ToByteArray();
byte[] order = { 15, 14, 13, 12, 11, 10, 9, 8, 6, 7, 4, 5, 0, 1, 2, 3 };
for (int i = 0; i < 16; i++)
{
if (bytes[order[i]] == byte.MaxValue)
{
bytes[order[i]] = 0;
}
else
{
bytes[order[i]]++;
return new Guid(bytes);
}
}
throw new OverflowException("Guid.Increment failed.");
}
}
}
using MyConsoleOne.BAL;
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace MyConsoleOne
{
class Program
{
static void Main(string[] args)
{
GuidStore gs = new GuidStore(Guid.NewGuid());
for (int i = 0; i < 1000; i++)
{
Console.WriteLine(i);
Dictionary<Guid, int> guids = new Dictionary<Guid, int>();
Parallel.For(0, 1000, j =>
{
Guid? currentGuid = gs.GetNextGuid();
guids.Add(currentGuid.Value, j);
Console.WriteLine(currentGuid);
}); // Parallel.For
}
Console.WriteLine("Press ENTER to Exit");
Console.ReadLine();
}
}
}
我的代码是:
的组合因为我得到&#34;为什么不使用Guid.NewGuid&#34;问题,我在这里提供原因:
我有一个父进程,它有一个由Guid.NewGuid()创建的uniqueidentifier。我将此称为&#34;父母guid&#34;。该父进程将创建N个文件。如果我是从头开始写作,我只会附加&#34; N&#34;在文件名的末尾。因此,如果父母Guid是&#34; 11111111-1111-1111-1111-111111111111&#34;例如,我会写文件
"11111111-1111-1111-1111-111111111111_1.txt"
"11111111-1111-1111-1111-111111111111_2.txt"
"11111111-1111-1111-1111-111111111111_3.txt"
等,但是,通过现有的&#34;合同&#34;与客户端:::文件名必须有一个(唯一的)Guid,而不是#34; N&#34; (1,2,等等)文件名中的值(这个&#34;合同&#34;已存在多年,所以它几乎是一成不变的)。通过这里列出的功能,我可以保留&#34;合同&#34;,但文件名与&#34; parent&#34;松散地联系在一起。 Guid(父母再次由Guid.NewGuid()生成)。碰撞 NOT 是文件名的问题(它们被放在一个不同的文件夹下,用于&#39;进程&#39;执行)。碰撞是&#34;父母的问题&#34; GUID。但同样,这已经用Guid.NewGuid处理了。
因此,使用&#34; 11111111-1111-1111-1111-111111111111&#34;的起始Guid,我将能够编写如下文件名:
OTHERSTUFF_111111111-1111-1111-1111-111111111112_MORESTUFF.txt
OTHERSTUFF_111111111-1111-1111-1111-111111111113_MORESTUFF.txt
OTHERSTUFF_111111111-1111-1111-1111-111111111114_MORESTUFF.txt
OTHERSTUFF_111111111-1111-1111-1111-111111111115_MORESTUFF.txt
OTHERSTUFF_111111111-1111-1111-1111-111111111116_MORESTUFF.txt
OTHERSTUFF_111111111-1111-1111-1111-111111111117_MORESTUFF.txt
OTHERSTUFF_111111111-1111-1111-1111-111111111118_MORESTUFF.txt
OTHERSTUFF_111111111-1111-1111-1111-111111111119_MORESTUFF.txt
OTHERSTUFF_111111111-1111-1111-1111-11111111111a_MORESTUFF.txt
OTHERSTUFF_111111111-1111-1111-1111-11111111111b_MORESTUFF.txt
所以在我上面的例子中,&#34;父母guid&#34;用#34; this.StartingGuid&#34; ....然后我得到&#34;递增&#34;指导原则。
还有。我可以编写更好的单元测试,因为现在我会提前知道文件名。
APPEND:
最终版代码:
public class GuidStore
{
private static object objectlock = new object();
private static int[] byteOrder = { 15, 14, 13, 12, 11, 10, 9, 8, 6, 7, 4, 5, 0, 1, 2, 3 };
private Guid StartingGuid { get; set; }
private Guid? LastGuidHolder { get; set; }
public GuidStore(Guid startingGuid)
{
this.StartingGuid = startingGuid;
}
public Guid GetNextGuid()
{
return this.GetNextGuid(0);
}
public Guid GetNextGuid(int firstGuidOffSet)
{
lock (objectlock)
{
if (this.LastGuidHolder.HasValue)
{
this.LastGuidHolder = Increment(this.LastGuidHolder.Value);
}
else
{
this.LastGuidHolder = Increment(this.StartingGuid);
for (int i = 0; i < firstGuidOffSet; i++)
{
this.LastGuidHolder = Increment(this.LastGuidHolder.Value);
}
}
return this.LastGuidHolder.Value;
}
}
private static Guid Increment(Guid guid)
{
var bytes = guid.ToByteArray();
var canIncrement = byteOrder.Any(i => ++bytes[i] != 0);
return new Guid(canIncrement ? bytes : new byte[16]);
}
}
和UnitTests:
public class GuidStoreUnitTests
{
[TestMethod]
public void GetNextGuidSimpleTest()
{
Guid startingGuid = new Guid("11111111-1111-1111-1111-111111111111");
GuidStore gs = new GuidStore(startingGuid);
List<Guid> guids = new List<Guid>();
const int GuidCount = 10;
for (int i = 0; i < GuidCount; i++)
{
guids.Add(gs.GetNextGuid());
}
Assert.IsNotNull(guids);
Assert.AreEqual(GuidCount, guids.Count);
Assert.IsNotNull(guids.FirstOrDefault(g => g == new Guid("11111111-1111-1111-1111-111111111112")));
Assert.IsNotNull(guids.FirstOrDefault(g => g == new Guid("11111111-1111-1111-1111-111111111113")));
Assert.IsNotNull(guids.FirstOrDefault(g => g == new Guid("11111111-1111-1111-1111-111111111114")));
Assert.IsNotNull(guids.FirstOrDefault(g => g == new Guid("11111111-1111-1111-1111-111111111115")));
Assert.IsNotNull(guids.FirstOrDefault(g => g == new Guid("11111111-1111-1111-1111-111111111116")));
Assert.IsNotNull(guids.FirstOrDefault(g => g == new Guid("11111111-1111-1111-1111-111111111117")));
Assert.IsNotNull(guids.FirstOrDefault(g => g == new Guid("11111111-1111-1111-1111-111111111118")));
Assert.IsNotNull(guids.FirstOrDefault(g => g == new Guid("11111111-1111-1111-1111-111111111119")));
Assert.IsNotNull(guids.FirstOrDefault(g => g == new Guid("11111111-1111-1111-1111-11111111111a")));
Assert.IsNotNull(guids.FirstOrDefault(g => g == new Guid("11111111-1111-1111-1111-11111111111b")));
}
[TestMethod]
public void GetNextGuidWithOffsetSimpleTest()
{
Guid startingGuid = new Guid("11111111-1111-1111-1111-111111111111");
GuidStore gs = new GuidStore(startingGuid);
List<Guid> guids = new List<Guid>();
const int OffSet = 10;
guids.Add(gs.GetNextGuid(OffSet));
Assert.IsNotNull(guids);
Assert.AreEqual(1, guids.Count);
Assert.IsNotNull(guids.FirstOrDefault(g => g == new Guid("11111111-1111-1111-1111-11111111111c")));
}
[TestMethod]
public void GetNextGuidMaxRolloverTest()
{
Guid startingGuid = new Guid("ffffffff-ffff-ffff-ffff-ffffffffffff");
GuidStore gs = new GuidStore(startingGuid);
List<Guid> guids = new List<Guid>();
const int OffSet = 10;
guids.Add(gs.GetNextGuid(OffSet));
Assert.IsNotNull(guids);
Assert.AreEqual(1, guids.Count);
Assert.IsNotNull(guids.FirstOrDefault(g => g == Guid.Empty));
}
[TestMethod]
public void GetNextGuidThreadSafeTest()
{
Guid startingGuid = Guid.NewGuid();
GuidStore gs = new GuidStore(startingGuid);
/* The "key" of the ConcurrentDictionary must be unique, so this will catch any duplicates */
ConcurrentDictionary<Guid, int> guids = new ConcurrentDictionary<Guid, int>();
Parallel.For(
0,
1000,
j =>
{
Guid currentGuid = gs.GetNextGuid();
if (!guids.TryAdd(currentGuid, j))
{
throw new ArgumentOutOfRangeException("GuidStore.GetNextGuid ThreadSafe Test Failed");
}
}); // Parallel.For
}
[TestMethod]
public void GetNextGuidTwoRunsProduceSameResultsTest()
{
Guid startingGuid = Guid.NewGuid();
GuidStore gsOne = new GuidStore(startingGuid);
/* The "key" of the ConcurrentDictionary must be unique, so this will catch any duplicates */
ConcurrentDictionary<Guid, int> setOneGuids = new ConcurrentDictionary<Guid, int>();
Parallel.For(
0,
1000,
j =>
{
Guid currentGuid = gsOne.GetNextGuid();
if (!setOneGuids.TryAdd(currentGuid, j))
{
throw new ArgumentOutOfRangeException("GuidStore.GetNextGuid ThreadSafe Test Failed");
}
}); // Parallel.For
gsOne = null;
GuidStore gsTwo = new GuidStore(startingGuid);
/* The "key" of the ConcurrentDictionary must be unique, so this will catch any duplicates */
ConcurrentDictionary<Guid, int> setTwoGuids = new ConcurrentDictionary<Guid, int>();
Parallel.For(
0,
1000,
j =>
{
Guid currentGuid = gsTwo.GetNextGuid();
if (!setTwoGuids.TryAdd(currentGuid, j))
{
throw new ArgumentOutOfRangeException("GuidStore.GetNextGuid ThreadSafe Test Failed");
}
}); // Parallel.For
bool equal = setOneGuids.Select(g => g.Key).OrderBy(i => i).SequenceEqual(
setTwoGuids.Select(g => g.Key).OrderBy(i => i), new GuidComparer<Guid>());
Assert.IsTrue(equal);
}
}
internal class GuidComparer<Guid> : IEqualityComparer<Guid>
{
public bool Equals(Guid x, Guid y)
{
return x.Equals(y);
}
public int GetHashCode(Guid obj)
{
return 0;
}
}
答案 0 :(得分:6)
这里有两个问题:
Dictionary.Add()
不是线程安全的。请改用ConcurrentDictionary.TryAdd()
。GetNextGuid()
的实现存在竞争条件,因为您正在锁定之外返回this.LastGuidHolder
,因此可以在返回之前由另一个线程进行修改。一个明显的解决方案是在锁内移动返回:
public Guid? GetNextGuid()
{
lock (objectlock)
{
if (this.LastGuidHolder.HasValue)
{
this.LastGuidHolder = Increment(this.LastGuidHolder.Value);
}
else
{
this.LastGuidHolder = Increment(this.StartingGuid);
}
return this.LastGuidHolder;
}
}
但是,我会将返回类型更改为Guid
- 它似乎没有任何目的可以返回Guid?
- 这应该隐藏在里面的东西班级:
public Guid GetNextGuid()
{
lock (objectlock)
{
if (this.LastGuidHolder.HasValue)
{
this.LastGuidHolder = Increment(this.LastGuidHolder.Value);
}
else
{
this.LastGuidHolder = Increment(this.StartingGuid);
}
return this.LastGuidHolder.Value;
}
}
以下是使用ConcurrentDictionary
的测试方法的一个版本:
static void Main(string[] args)
{
GuidStore gs = new GuidStore(Guid.NewGuid());
for (int i = 0; i < 1000; i++)
{
Console.WriteLine(i);
ConcurrentDictionary<Guid, int> guids = new ConcurrentDictionary<Guid, int>();
Parallel.For(0, 1000, j =>
{
Guid currentGuid = gs.GetNextGuid();
if (!guids.TryAdd(currentGuid, j))
{
Console.WriteLine("Duplicate found!");
}
}); // Parallel.For
}
Console.WriteLine("Press ENTER to Exit");
Console.ReadLine();
}
说了这么多,我无法理解你为什么不只是使用Guid.NewGuid()
...