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Revert "Simplify DoubleQueue to eliminate redundant sempahore work." 

This reverts commit 52b7b40034.

Got the semantics wrong - the sempahore is required so that the blocking thread waits for a signal.
0.8.0.3
Justin Clark-Casey (justincc) 2014-03-19 01:40:56 +00:00
parent 52b7b40034
commit cf97535d9e
1 changed files with 27 additions and 7 deletions
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32
OpenSim/Framework/Util.cs
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@ -2296,6 +2296,11 @@ namespace OpenSim.Framework
private Queue<T> m_highQueue = new Queue<T>();
private object m_syncRoot = new object();
private Semaphore m_s = new Semaphore(0, 1);
public DoubleQueue()
{
}
public virtual int Count
{
@ -2324,7 +2329,11 @@ namespace OpenSim.Framework
private void Enqueue(Queue<T> q, T data)
{
lock (m_syncRoot)
{
q.Enqueue(data);
m_s.WaitOne(0);
m_s.Release();
}
}
public virtual T Dequeue()
@ -2354,28 +2363,39 @@ namespace OpenSim.Framework
public bool Dequeue(TimeSpan wait, ref T res)
{
if (!Monitor.TryEnter(m_syncRoot, wait))
if (!m_s.WaitOne(wait))
return false;
try
lock (m_syncRoot)
{
if (m_highQueue.Count > 0)
res = m_highQueue.Dequeue();
else if (m_lowQueue.Count > 0)
res = m_lowQueue.Dequeue();
if (m_highQueue.Count == 0 && m_lowQueue.Count == 0)
return true;
try
{
m_s.Release();
}
catch
{
}
return true;
}
finally
{
Monitor.Exit(m_syncRoot);
}
}
public virtual void Clear()
{
lock (m_syncRoot)
{
// Make sure sem count is 0
m_s.WaitOne(0);
m_lowQueue.Clear();
m_highQueue.Clear();
}