假设以下代码在我的应用程序中以类似的方式使用:
//-------------------------------------
void UseAllResources ()
{
bool bSuccess1 = false;
bool bSuccess2 = false;
try
{
bSuccess1 = Monitor::TryEnter (oResource1, msc_iTimeoutMonitor);
if (!bSuccess1) return;
bSuccess2 = Monitor::TryEnter (oResource2, msc_iTimeoutMonitor);
if (!bSuccess2) return;
// work on oResource1 and oResource2
} finally {
if (bSuccess2)
Monitor::Exit (oResource2);
if (bSuccess1)
Monitor::Exit (oResource1);
}
}
//-------------------------------------
void UseResource1 ()
{
bool bSuccess = false;
try {
bSuccess = Monitor::TryEnter (oResource1, msc_iTimeoutMonitor);
if (!bSuccess) return;
// work on oResource1
} finally {
if (bSuccess) Monitor::Exit (oResource1);
}
}
//-------------------------------------
void UseResource2 ()
{
same like UseResource1(), but using oResource2
}
这些函数可能随时被不同的线程调用。
可能会发生
(超时为500ms)
@t=0ms,线程B正在执行UseResource2()
,需要400ms,
@t=100ms,线程 Z 正在调用 UseAllResources(),获取了 oResource1 上的锁,必须等待 oResource2 上的锁,
@t=200ms,线程 A 正在调用 UseResource1() 并且必须等待 oResource1 上的锁,它被线程 Z 占用,
@t=400ms,线程B完成,线程Z获取oResource2
上的锁并开始工作,需要400ms,
@t=700ms,线程 A 超时,尽管它只需要 50 毫秒并且可以在线程 Z 仍在等待时工作。
我宁愿线程 Z 失败,如果有的话,因为超时应该是所有锁的总值。
我可以同时开始获取多个锁吗?
最佳答案
一个解决方案可能是使用 ReaderWriterLockSlim 类。下面的代码在构造函数中包装了一个函数(你要做的工作)。或者,您可以将该函数移至 DoWork 方法以更改您访问资源的方式。
LockedResource 实现
class LockedResource
{
public delegate void RefAction();
ReaderWriterLockSlim resourceLock;
public LockedResource()
{
//Warning: SupportsRecursion is risky, you should remove support for recursive whenever possible
resourceLock = new ReaderWriterLockSlim(LockRecursionPolicy.SupportsRecursion);
}
public bool DoWork(RefAction work, string threadname, int timeout = -1)
{
try
{
if (resourceLock.TryEnterWriteLock(timeout))
{
if (work != null)
{
work();
}
}
else
{
Console.WriteLine("Lock time out on thread {0}", threadname);
}
}
finally
{
Console.WriteLine("{0} releasing resource", threadname);
if(resourceLock.IsWriteLockHeld)
{
resourceLock.ExitWriteLock();
}
}
return false;
}
}
示例用法
static void Main(string[] args)
{
object oResouce1 = "-";
object oResouce2 = "-";
LockedResource lock1 = new LockedResource();
LockedResource lock2 = new LockedResource();
//the event wait handles is not required, only used to block thread so that resource values can be printed out at the end of the program
var h1 = new EventWaitHandle(false, EventResetMode.ManualReset);
var h2 = new EventWaitHandle(false, EventResetMode.ManualReset);
var h3 = new EventWaitHandle(false, EventResetMode.ManualReset);
WaitHandle[] waitHandles = { h1, h2, h3 };
var t1 = new Thread(() =>
{
lock1.DoWork(() =>
{
oResouce1 = "1";
Console.WriteLine("Resource 1 set to 1");
},"T1");
h1.Set();
});
var t2 = new Thread(() =>
{
lock2.DoWork(() =>
{
oResouce2 = "2";
Console.WriteLine("Resource 2 set to 2");
Thread.Sleep(10000);
}, "T2");
h2.Set();
});
var t3 = new Thread(() =>
{
lock1.DoWork(() =>
{
lock2.DoWork(() =>
{
oResouce1 = "3";
Console.WriteLine("Resource 1 set to 3");
oResouce2 = "3";
Console.WriteLine("Resource 2 set to 3");
}, "T3", 1000);
h3.Set();
}, "T3");
});
t1.Start();
t2.Start();
t3.Start();
WaitHandle.WaitAll(waitHandles);
Console.WriteLine("Resource 1 is {0}", oResouce1);
Console.WriteLine("Resource 2 is {0}", oResouce2);
Console.ReadLine();
}
输出
Resource 1 set to 1
Resource 2 set to 2
T1 releasing resource
Lock time out on thread T3
T3 releasing resource
T3 releasing resource
T2 releasing resource
Resource 1 is 1
Resource 2 is 2
关于c# - 同时获取多个线程同步锁,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37520451/