一般介绍是,操作都是先wait, 再signal,但是如果先signal再wait,后果是怎样,找不到结论,一直怀疑这里有问题。自己做实验,发觉也没有问题。
我开启了两个线程,一个计时器,以下就是我的关键代码:
TInt CTestMultiThreadAppUi::ThreadProc1(void* pParam)
{
CTestMultiThreadAppUi* pApp = (CTestMultiThreadAppUi*)pParam;
while(!pApp->iStop)
{
if(pApp->iLocked)
{
pApp->iLocked = false;
pApp->m_cSemaphore.Wait();
}
User::After(10000);
}
return 0;
}
TInt CTestMultiThreadAppUi::ThreadProc2(void* pParam)
{
CTestMultiThreadAppUi* pApp = (CTestMultiThreadAppUi*)pParam;
while(!pApp->iStop)
{
if(pApp->iUnLocked)
{
pApp->iUnLocked = false;
pApp->m_cSemaphore.Signal();
}
User::After(10000);
}
return 0;
}
TInt CTestMultiThreadAppUi::TimerLoop(void* pParam)
{
CTestMultiThreadAppUi* pApp = (CTestMultiThreadAppUi*)pParam;
pApp->iCount ++;
if(pApp->iCount == 1000)
pApp->iLocked = true;
if(pApp->iCount == 100)
pApp->iUnLocked = true;
return 0;
}
------解决方案--------------------------------------------------------
RSemaphore 创建后的初始值决定的,如果值少于等于0时,线程阻塞,初始值为1 .
那么先wait, 再signal,调用WAIT的线程阻塞,反过来先signal再wait,协同工作的线程都不会阻塞,
可能出现冲突