Potencial problems in SemaphoreSlimExtensions class

[TomatoDinosaurEgg]

I see the source code of SemaphoreSlimExtensions class. I have a question on the method Task<IDisposable> LockAsync(this SemaphoreSlim semaphoreSlim, int millisecondsTimeout)

The general way we use Task<bool> WaitAsync(int millisecondsTimeout) is like below:

SemaphoreSlim mutex = new SemaphoreSlim(1, 1);
if(await mutex.WaitAsync(6000))
{
     //code to be locked.
}
else
{
     throw new Exception("Wait for mutex time out");
}

If we use LockAsync method in SemaphoreSlimExtensions class, like below:

SemaphoreSlim mutex = new SemaphoreSlim(1, 1);
using (await mutex.LockAsync(6000))
{
      //code to be locked.
}

I found some problems, if wait the mutex for 6s, and failed to get the mutex, it will still execute the code need to be locked. And there is no way to check whether entered the mutex successfully or not during the specific "TimeOut".

I code some demos to test the problem.

var mutex = new SemaphoreSlim(1, 1);
int count = 0;
for(int i = 0; i < 10; i++)
{
        Task.Run(async () =>
        {
             if (await mutex.WaitAsync(1000))
             {
                   await Task.Delay(200);
                  Console.WriteLine(count++);
              }
              else
              {
                   Console.WriteLine("wait for mutex time out");
              }
              mutex.Release();

        });
}

The result is expected

However, when I use LockAsync(), the result is unexpected.

var mutex = new SemaphoreSlim(1, 1);
int count = 0;
for(int i = 0; i < 10; i++)
{
        Task.Run(async () =>
        {
             using (await mutex.LockAsync(1000))
            {
                await Task.Delay(200);
                Console.WriteLine(count++);
            }
        });
}

Duplicate number printed, the code was not locked as expected.

Am I misunderstand something, or it's a bug?

[maliming]

hi

I will confirm that. Thanks for feedback.

[maliming]

hi
#16645