在窗口的创建过程中
AfxBeginThread(ThreadProc1, NULL, THREAD_PRIORITY_TIME_CRITICAL);
UINT ThreadProc1(LPVOID pParam)
{
while(1)
{
}
}
会有什么结果?
如果我再开一个呢?
AfxBeginThread(ThreadProc2, NULL, THREAD_PRIORITY_TIME_CRITICAL);
UINT ThreadProc2(LPVOID pParam)
{
for (;;)
{
}
}
各位, 有兴趣的可以试验一下?
AfxBeginThread(ThreadProc1, NULL, THREAD_PRIORITY_TIME_CRITICAL);
UINT ThreadProc1(LPVOID pParam)
{
while(1)
{
}
}
会有什么结果?
如果我再开一个呢?
AfxBeginThread(ThreadProc2, NULL, THREAD_PRIORITY_TIME_CRITICAL);
UINT ThreadProc2(LPVOID pParam)
{
for (;;)
{
}
}
各位, 有兴趣的可以试验一下?
UINT ThreadProc1(LPVOID pParam)
{
UINT nCount = 0;
while(1)
{
Sleep(1000);
cout<<"from Thread 1:"<<nCount++<<endl;
}
} AfxBeginThread(ThreadProc2, NULL, THREAD_PRIORITY_TIME_CRITICAL);
UINT ThreadProc2(LPVOID pParam)
{
UINT nCount = 0;
while(1)
{
Sleep(1000);
cout<<"from Thread 2:"<<nCount++<<endl;
}
} ......AfxBeginThread(ThreadProc1000, NULL, THREAD_PRIORITY_TIME_CRITICAL);
UINT ThreadProc1000(LPVOID pParam)
{
UINT nCount = 0;
while(1)
{
Sleep(1000);
cout<<"from Thread 1000:"<<nCount++<<endl;
}
} 也就是说开启1000个这样的线程,事实证明,这1000个线程执行的时间是不相同的,也就是有些线程执行较多,有些则很少,如何使它们的执行机会均等呢?