在 C# 中使用全局互斥锁的好模式是什么？

What is a good pattern for using a Global Mutex in C#?

提问by Sam Saffron

The Mutex class is very misunderstood, and Global mutexes even more so.

Mutex 类很容易被误解，而 Global mutex 更是如此。

What is good, safe pattern to use when creating Global mutexes?

创建全局互斥锁时使用什么好的、安全的模式？

One that will work

一个会起作用的

• Regardless of the locale my machine is in
• Is guaranteed to release the mutex properly
• Optionally does not hang forever if the mutex is not acquired
• Deals with cases where other processes abandon the mutex

• 无论我的机器所处的语言环境如何
• 保证正确释放互斥锁
• 如果未获取互斥锁，则可选地不会永远挂起
• 处理其他进程放弃互斥锁的情况

采纳答案by Sam Saffron

I want to make sure this is out there, because it's so hard to get right:

我想确保这是在那里，因为很难做到正确：

using System.Runtime.InteropServices;   //GuidAttribute
using System.Reflection;                //Assembly
using System.Threading;                 //Mutex
using System.Security.AccessControl;    //MutexAccessRule
using System.Security.Principal;        //SecurityIdentifier

static void Main(string[] args)
{
    // get application GUID as defined in AssemblyInfo.cs
    string appGuid =
        ((GuidAttribute)Assembly.GetExecutingAssembly().
            GetCustomAttributes(typeof(GuidAttribute), false).
                GetValue(0)).Value.ToString();

    // unique id for global mutex - Global prefix means it is global to the machine
    string mutexId = string.Format( "Global\{{{0}}}", appGuid );

    // Need a place to store a return value in Mutex() constructor call
    bool createdNew;

    // edited by Jeremy Wiebe to add example of setting up security for multi-user usage
    // edited by 'Marc' to work also on localized systems (don't use just "Everyone") 
    var allowEveryoneRule =
        new MutexAccessRule( new SecurityIdentifier( WellKnownSidType.WorldSid
                                                   , null)
                           , MutexRights.FullControl
                           , AccessControlType.Allow
                           );
    var securitySettings = new MutexSecurity();
    securitySettings.AddAccessRule(allowEveryoneRule);

   // edited by MasonGZhwiti to prevent race condition on security settings via VanNguyen
    using (var mutex = new Mutex(false, mutexId, out createdNew, securitySettings))
    {
        // edited by acidzombie24
        var hasHandle = false;
        try
        {
            try
            {
                // note, you may want to time out here instead of waiting forever
                // edited by acidzombie24
                // mutex.WaitOne(Timeout.Infinite, false);
                hasHandle = mutex.WaitOne(5000, false);
                if (hasHandle == false)
                    throw new TimeoutException("Timeout waiting for exclusive access");
            }
            catch (AbandonedMutexException)
            {
                // Log the fact that the mutex was abandoned in another process,
                // it will still get acquired
                hasHandle = true;
            }

            // Perform your work here.
        }
        finally
        {
            // edited by acidzombie24, added if statement
            if(hasHandle)
                mutex.ReleaseMutex();
        }
    }
}

回答by Liam

This example will exit after 5 seconds if another instance is already running.

如果另一个实例已在运行，则此示例将在 5 秒后退出。

// unique id for global mutex - Global prefix means it is global to the machine
const string mutex_id = "Global\{B1E7934A-F688-417f-8FCB-65C3985E9E27}";

static void Main(string[] args)
{

    using (var mutex = new Mutex(false, mutex_id))
    {
        try
        {
            try
            {
                if (!mutex.WaitOne(TimeSpan.FromSeconds(5), false))
                {
                    Console.WriteLine("Another instance of this program is running");
                    Environment.Exit(0);
                }
            }
            catch (AbandonedMutexException)
            {
                // Log the fact the mutex was abandoned in another process, it will still get aquired
            }

            // Perform your work here.
        }
        finally
        {
            mutex.ReleaseMutex();
        }
    }
}

回答by deepee1

Using the accepted answer I create a helper class so you could use it in a similar way you would use the Lock statement. Just thought I'd share.

使用接受的答案，我创建了一个辅助类，以便您可以像使用 Lock 语句一样使用它。只是想我会分享。

Use:

用：

using (new SingleGlobalInstance(1000)) //1000ms timeout on global lock
{
    //Only 1 of these runs at a time
    RunSomeStuff();
}

And the helper class:

和助手类：

class SingleGlobalInstance : IDisposable
{
    //edit by user "jitbit" - renamed private fields to "_"
    public bool _hasHandle = false;
    Mutex _mutex;

    private void InitMutex()
    {
        string appGuid = ((GuidAttribute)Assembly.GetExecutingAssembly().GetCustomAttributes(typeof(GuidAttribute), false).GetValue(0)).Value;
        string mutexId = string.Format("Global\{{{0}}}", appGuid);
        _mutex = new Mutex(false, mutexId);

        var allowEveryoneRule = new MutexAccessRule(new SecurityIdentifier(WellKnownSidType.WorldSid, null), MutexRights.FullControl, AccessControlType.Allow);
        var securitySettings = new MutexSecurity();
        securitySettings.AddAccessRule(allowEveryoneRule);
        _mutex.SetAccessControl(securitySettings);
    }

    public SingleGlobalInstance(int timeOut)
    {
        InitMutex();
        try
        {
            if(timeOut < 0)
                _hasHandle = _mutex.WaitOne(Timeout.Infinite, false);
            else
                _hasHandle = _mutex.WaitOne(timeOut, false);

            if (_hasHandle == false)
                throw new TimeoutException("Timeout waiting for exclusive access on SingleInstance");
        }
        catch (AbandonedMutexException)
        {
            _hasHandle = true;
        }
    }


    public void Dispose()
    {
        if (_mutex != null)
        {
            if (_hasHandle)
                _mutex.ReleaseMutex();
            _mutex.Close();
        }
    }
}

回答by sol

Neither Mutex nor WinApi CreateMutex() works for me.

Mutex 和 WinApi CreateMutex() 都不适合我。

An alternate solution:

替代解决方案：

static class Program
{
    [STAThread]
    static void Main()
    {
        if (SingleApplicationDetector.IsRunning()) {
            return;
        }

        Application.Run(new MainForm());

        SingleApplicationDetector.Close();
    }
}

And the SingleApplicationDetector:

和SingleApplicationDetector：

using System;
using System.Reflection;
using System.Runtime.InteropServices;
using System.Security.AccessControl;
using System.Threading;

public static class SingleApplicationDetector
{
    public static bool IsRunning()
    {
        string guid = ((GuidAttribute)Assembly.GetExecutingAssembly().GetCustomAttributes(typeof(GuidAttribute), false).GetValue(0)).Value.ToString();
        var semaphoreName = @"Global\" + guid;
        try {
            __semaphore = Semaphore.OpenExisting(semaphoreName, SemaphoreRights.Synchronize);

            Close();
            return true;
        }
        catch (Exception ex) {
            __semaphore = new Semaphore(0, 1, semaphoreName);
            return false;
        }
    }

    public static void Close()
    {
        if (__semaphore != null) {
            __semaphore.Close();
            __semaphore = null;
        }
    }

    private static Semaphore __semaphore;
}

Reason to use Semaphore instead of Mutex:

使用 Semaphore 而不是 Mutex 的原因：

The Mutex class enforces thread identity, so a mutex can be released only by the thread that acquired it. By contrast, the Semaphore class does not enforce thread identity.

<< System.Threading.Mutex

Mutex 类强制执行线程标识，因此只有获得它的线程才能释放互斥锁。相比之下， Semaphore 类不强制线程标识。

<< System.Threading.Mutex

Ref: Semaphore.OpenExisting()

参考：Semaphore.OpenExisting()

回答by Van Nguyen

There is a race condition in the accepted answer when 2 processes running under 2 different users trying to initialize the mutex at the same time. After the first process initializes the mutex, if the second process tries to initialize the mutex before the first process sets the access rule to everyone, an unauthorized exception will be thrown by the second process.

当在 2 个不同用户下运行的 2 个进程试图同时初始化互斥锁时，接受的答案中存在竞争条件。在第一个进程初始化互斥锁之后，如果第二个进程在第一个进程对每个人设置访问规则之前尝试初始化互斥锁，则第二个进程将抛出未经授权的异常。

See below for corrected answer:

请参阅下面的更正答案：

using System.Runtime.InteropServices;   //GuidAttribute
using System.Reflection;                //Assembly
using System.Threading;                 //Mutex
using System.Security.AccessControl;    //MutexAccessRule
using System.Security.Principal;        //SecurityIdentifier

static void Main(string[] args)
{
    // get application GUID as defined in AssemblyInfo.cs
    string appGuid = ((GuidAttribute)Assembly.GetExecutingAssembly().GetCustomAttributes(typeof(GuidAttribute), false).GetValue(0)).Value.ToString();

    // unique id for global mutex - Global prefix means it is global to the machine
    string mutexId = string.Format( "Global\{{{0}}}", appGuid );

    bool createdNew;
        // edited by Jeremy Wiebe to add example of setting up security for multi-user usage
        // edited by 'Marc' to work also on localized systems (don't use just "Everyone") 
        var allowEveryoneRule = new MutexAccessRule(new SecurityIdentifier(WellKnownSidType.WorldSid, null), MutexRights.FullControl, AccessControlType.Allow);
        var securitySettings = new MutexSecurity();
        securitySettings.AddAccessRule(allowEveryoneRule);

        using (var mutex = new Mutex(false, mutexId, out createdNew, securitySettings))
        {

        // edited by acidzombie24
        var hasHandle = false;
        try
        {
            try
            {
                // note, you may want to time out here instead of waiting forever
                // edited by acidzombie24
                // mutex.WaitOne(Timeout.Infinite, false);
                hasHandle = mutex.WaitOne(5000, false);
                if (hasHandle == false)
                    throw new TimeoutException("Timeout waiting for exclusive access");
            }
            catch (AbandonedMutexException)
            {
                // Log the fact the mutex was abandoned in another process, it will still get aquired
                hasHandle = true;
            }

            // Perform your work here.
        }
        finally
        {
            // edited by acidzombie24, added if statemnet
            if(hasHandle)
                mutex.ReleaseMutex();
        }
    }
}

回答by TruthSeeker

Sometimes learning by example helps the most. Run this console application in three different console windows. You'll see that the application you ran first acquires the mutex first, while the other two are waiting their turn. Then press enter in the first application, you'll see that application 2 now continues running by acquiring the mutex, however application 3 is waiting its turn. After you press enter in application 2 you'll see that application 3 continues. This illustrates the concept of a mutex protecting a section of code to be executed only by one thread (in this case a process) like writing to a file as an example.

有时，以身作则最有帮助。在三个不同的控制台窗口中运行此控制台应用程序。您将看到您首先运行的应用程序首先获取互斥锁，而其他两个正在等待轮到它们。然后在第一个应用程序中按 Enter，您将看到应用程序 2 现在通过获取互斥量继续运行，但是应用程序 3 正在等待轮到它。在应用程序 2 中按 Enter 后，您将看到应用程序 3 继续。这说明了互斥锁的概念，保护一段代码只能由一个线程（在本例中为进程）执行，例如写入文件。

using System;
using System.Threading;

namespace MutexExample
{
    class Program
    {
        static Mutex m = new Mutex(false, "myMutex");//create a new NAMED mutex, DO NOT OWN IT
        static void Main(string[] args)
        {
            Console.WriteLine("Waiting to acquire Mutex");
            m.WaitOne(); //ask to own the mutex, you'll be queued until it is released
            Console.WriteLine("Mutex acquired.\nPress enter to release Mutex");
            Console.ReadLine();
            m.ReleaseMutex();//release the mutex so other processes can use it
        }
    }
}

回答by Eric Ouellet

A global Mutex is not only to ensure to have only one instance of an application. I personally prefer using Microsoft.VisualBasic to ensure single instance application like described in What is the correct way to create a single-instance WPF application?(Dale Ragan answer)... I found that's easier to pass arguments received on new application startup to the initial single instance application.

一个全局的 Mutex 不仅仅是保证一个应用程序只有一个实例。我个人更喜欢使用 Microsoft.VisualBasic 来确保单实例应用程序，如创建单实例 WPF 应用程序的正确方法是什么？（Dale Ragan 回答）...我发现将新应用程序启动时收到的参数传递给初始单实例应用程序更容易。

But regarding some previous code in this thread, I would prefer to not create a Mutex each time I want to have a lock on it. It could be fine for a single instance application but in other usage it appears to me has overkill.

但是对于这个线程中的一些以前的代码，我不想每次想要锁定它时都创建一个互斥锁。对于单个实例应用程序来说可能没问题，但在其他用途​​中，在我看来它有点矫枉过正。

That's why I suggest this implementation instead:

这就是为什么我建议使用此实现：

Usage:

用法：

static MutexGlobal _globalMutex = null;
static MutexGlobal GlobalMutexAccessEMTP
{
    get
    {
        if (_globalMutex == null)
        {
            _globalMutex = new MutexGlobal();
        }
        return _globalMutex;
    }
}

using (GlobalMutexAccessEMTP.GetAwaiter())
{
    ...
}   

Mutex Global Wrapper:

互斥全局包装器：

using System;
using System.Reflection;
using System.Runtime.InteropServices;
using System.Security.AccessControl;
using System.Security.Principal;
using System.Threading;

namespace HQ.Util.General.Threading
{
    public class MutexGlobal : IDisposable
    {
        // ************************************************************************
        public string Name { get; private set; }
        internal Mutex Mutex { get; private set; }
        public int DefaultTimeOut { get; set; }
        public Func<int, bool> FuncTimeOutRetry { get; set; }

        // ************************************************************************
        public static MutexGlobal GetApplicationMutex(int defaultTimeOut = Timeout.Infinite)
        {
            return new MutexGlobal(defaultTimeOut, ((GuidAttribute)Assembly.GetExecutingAssembly().GetCustomAttributes(typeof(GuidAttribute), false).GetValue(0)).Value);
        }

        // ************************************************************************
        public MutexGlobal(int defaultTimeOut = Timeout.Infinite, string specificName = null)
        {
            try
            {
                if (string.IsNullOrEmpty(specificName))
                {
                    Name = Guid.NewGuid().ToString();
                }
                else
                {
                    Name = specificName;
                }

                Name = string.Format("Global\{{{0}}}", Name);

                DefaultTimeOut = defaultTimeOut;

                FuncTimeOutRetry = DefaultFuncTimeOutRetry;

                var allowEveryoneRule = new MutexAccessRule(new SecurityIdentifier(WellKnownSidType.WorldSid, null), MutexRights.FullControl, AccessControlType.Allow);
                var securitySettings = new MutexSecurity();
                securitySettings.AddAccessRule(allowEveryoneRule);

                Mutex = new Mutex(false, Name, out bool createdNew, securitySettings);

                if (Mutex == null)
                {
                    throw new Exception($"Unable to create mutex: {Name}");
                }
            }
            catch (Exception ex)
            {
                Log.Log.Instance.AddEntry(Log.LogType.LogException, $"Unable to create Mutex: {Name}", ex);
                throw;
            }
        }

        // ************************************************************************
        /// <summary>
        /// 
        /// </summary>
        /// <param name="timeOut"></param>
        /// <returns></returns>
        public MutexGlobalAwaiter GetAwaiter(int timeOut)
        {
            return new MutexGlobalAwaiter(this, timeOut);
        }

        // ************************************************************************
        /// <summary>
        /// 
        /// </summary>
        /// <param name="timeOut"></param>
        /// <returns></returns>
        public MutexGlobalAwaiter GetAwaiter()
        {
            return new MutexGlobalAwaiter(this, DefaultTimeOut);
        }

        // ************************************************************************
        /// <summary>
        /// This method could either throw any user specific exception or return 
        /// true to retry. Otherwise, retruning false will let the thread continue
        /// and you should verify the state of MutexGlobalAwaiter.HasTimedOut to 
        /// take proper action depending on timeout or not. 
        /// </summary>
        /// <param name="timeOutUsed"></param>
        /// <returns></returns>
        private bool DefaultFuncTimeOutRetry(int timeOutUsed)
        {
            // throw new TimeoutException($"Mutex {Name} timed out {timeOutUsed}.");

            Log.Log.Instance.AddEntry(Log.LogType.LogWarning, $"Mutex {Name} timeout: {timeOutUsed}.");
            return true; // retry
        }

        // ************************************************************************
        public void Dispose()
        {
            if (Mutex != null)
            {
                Mutex.ReleaseMutex();
                Mutex.Close();
            }
        }

        // ************************************************************************

    }
}

Awaiter

服务员

using System;

namespace HQ.Util.General.Threading
{
    public class MutexGlobalAwaiter : IDisposable
    {
        MutexGlobal _mutexGlobal = null;

        public bool HasTimedOut { get; set; } = false;

        internal MutexGlobalAwaiter(MutexGlobal mutexEx, int timeOut)
        {
            _mutexGlobal = mutexEx;

            do
            {
                HasTimedOut = !_mutexGlobal.Mutex.WaitOne(timeOut, false);
                if (! HasTimedOut) // Signal received
                {
                    return;
                }
            } while (_mutexGlobal.FuncTimeOutRetry(timeOut));
        }

        #region IDisposable Support
        private bool disposedValue = false; // To detect redundant calls

        protected virtual void Dispose(bool disposing)
        {
            if (!disposedValue)
            {
                if (disposing)
                {
                    _mutexGlobal.Mutex.ReleaseMutex();
                }

                // TODO: free unmanaged resources (unmanaged objects) and override a finalizer below.
                // TODO: set large fields to null.

                disposedValue = true;
            }
        }
        // TODO: override a finalizer only if Dispose(bool disposing) above has code to free unmanaged resources.
        // ~MutexExAwaiter()
        // {
        //   // Do not change this code. Put cleanup code in Dispose(bool disposing) above.
        //   Dispose(false);
        // }

        // This code added to correctly implement the disposable pattern.
        public void Dispose()
        {
            // Do not change this code. Put cleanup code in Dispose(bool disposing) above.
            Dispose(true);
            // TODO: uncomment the following line if the finalizer is overridden above.
            // GC.SuppressFinalize(this);
        }
        #endregion
    }
}

回答by Wouter

A solution (for WPF) without WaitOne because it can cause an AbandonedMutexException. This solution uses the Mutex constructor that returns the createdNew boolean to check if the mutex is already created. It also uses the GetType().GUID so renaming an executable doesn't allow multiple instances.

没有 WaitOne 的解决方案（用于 WPF），因为它可能导致 AbandonedMutexException。此解决方案使用 Mutex 构造函数返回 createdNew 布尔值来检查互斥锁是否已创建。它还使用 GetType().GUID，因此重命名可执行文件不允许多个实例。

Global vs local mutex see note in: https://docs.microsoft.com/en-us/dotnet/api/system.threading.mutex?view=netframework-4.8

全局与本地互斥体参见注释：https: //docs.microsoft.com/en-us/dotnet/api/system.threading.mutex?view=netframework-4.8

private Mutex mutex;
private bool mutexCreated;

public App()
{
    string mutexId = $"Global\{GetType().GUID}";
    mutex = new Mutex(true, mutexId, out mutexCreated);
}

protected override void OnStartup(StartupEventArgs e)
{
    base.OnStartup(e);
    if (!mutexCreated)
    {
        MessageBox.Show("Already started!");
        Shutdown();
    }
}

Because Mutex implements IDisposable it is released automatically but for completeness call dispose:

因为 Mutex 实现了 IDisposable，它会自动释放，但为了完整性调用 dispose：

protected override void OnExit(ExitEventArgs e)
{
    base.OnExit(e);
    mutex.Dispose();
}

Move everything into a base class and add the allowEveryoneRule from the accepted answer. Also added ReleaseMutex though it doesn't look like it's really needed because it is released automatically by the OS (what if the application crashes and never calls ReleaseMutex would you need to reboot?).

将所有内容移到基类中，并从接受的答案中添加 allowEveryoneRule。还添加了 ReleaseMutex 虽然它看起来并不像真的需要它，因为它是由操作系统自动释放的（如果应用程序崩溃并且从不调用 ReleaseMutex 会怎样，您需要重新启动吗？）。

public class SingleApplication : Application
{
    private Mutex mutex;
    private bool mutexCreated;

    public SingleApplication()
    {
        string mutexId = $"Global\{GetType().GUID}";

        MutexAccessRule allowEveryoneRule = new MutexAccessRule(
            new SecurityIdentifier(WellKnownSidType.WorldSid, null),
            MutexRights.FullControl, 
            AccessControlType.Allow);
        MutexSecurity securitySettings = new MutexSecurity();
        securitySettings.AddAccessRule(allowEveryoneRule);

        // initiallyOwned: true == false + mutex.WaitOne()
        mutex = new Mutex(initiallyOwned: true, mutexId, out mutexCreated, securitySettings);        }

    protected override void OnExit(ExitEventArgs e)
    {
        base.OnExit(e);
        if (mutexCreated)
        {
            try
            {
                mutex.ReleaseMutex();
            }
            catch (ApplicationException ex)
            {
                MessageBox.Show(ex.Message, ex.GetType().FullName, MessageBoxButton.OK, MessageBoxImage.Error);
            }
        }
        mutex.Dispose();
    }

    protected override void OnStartup(StartupEventArgs e)
    {
        base.OnStartup(e);
        if (!mutexCreated)
        {
            MessageBox.Show("Already started!");
            Shutdown();
        }
    }
}