c# Mutex 问题：从不同步的代码块访问对象同步方法

Question

发生概率：几个月来只发生了一两次，目前来看概率很小；
问题描述：我从外部线程去访问某个队列实例的MessageQueuePop有时会出现如下错误： 【ReleaseMutex报错，说明代码已经进入WaitOne，不知引起异常的原因。】
[2017-10-30 10:50:53,237] TD000002_baseThread -- DEBUG -- AutomationStation.Utils.MessageQueue [MessageQueuePop] -- System.ApplicationException: Object synchronization method was called from an unsynchronized block of code.
at System.Threading.Mutex.ReleaseMutex()
at AutomationStation.Utils.MessageQueue.MessageQueuePop()

 -------消息队列类的代码如下：-----------------------------------------

```using System;
using System.Threading;
using AutomationStation.Interface;

namespace AutomationStation.Utils
{
public class MessageQueue: IMessageQueue
{
private int count;
private int widx;
private int ridx;
private string[] cmd;
private Mutex mutex;
private log4net.ILog log4 = log4net.LogManager.GetLogger(System.Reflection.MethodBase.GetCurrentMethod().DeclaringType);

    public MessageQueue()
    {
        IMessageQueueInit();
    }
    public void IMessageQueueInit()
    {
        count = 0;
        widx = 0;
        ridx = 0;
        cmd = new string[100];
        mutex = new Mutex();
    }
    public void IMessageQueueInit(string blockName, string messageQueueName)
    {
        throw new NotSupportedException();
    }

    public void MessageQueuePush(string msg)
    {
        try
        {
            mutex.WaitOne();
            if (count >= 100)
            {
                mutex.ReleaseMutex();
                return;
            }
            cmd[widx] = msg;
            widx = (widx + 1) % 100;
            count++;
            mutex.ReleaseMutex();
        }
        catch (Exception ex) { log4.Debug(ex.ToString()); }
    }

    public string MessageQueuePop()
    {
        string msg = "";
        try
        {
            mutex.WaitOne();
            if (count <= 0)
            {
                mutex.ReleaseMutex();
                return msg;
            }
            msg = cmd[ridx];
            ridx = (ridx + 1) % 100;
            count--;
            mutex.ReleaseMutex();
        }
        catch (Exception ex) { log4.Debug(ex.ToString()); }
        return msg;
    }

    public void MessageQueueClear()
    {
        mutex.WaitOne();
        widx = 0;
        ridx = 0;
        count = 0;
        mutex.ReleaseMutex();
    }


}

}

Answer 1

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