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我在写一个基本的生产者/消费者线程代码。我得到了大部分的工作,但我有一个问题与兰特()。或者,也许我的问题比rand()更深,而rand只是冰山一角。我不想做任何太复杂的事情(无法运行或等待)。多线程生产者/消费者
我有一个全局的整数作为缓冲区。我允许用户输入大小和运行时间的限制。 我让计数器成为一个静态全局变量。 这是我的制片人:
DWORD WINAPI producer(LPVOID n)
{
cout << "\nPRODUCER:The producer is running right now" << endl;
int size = (int)n;
int num = rand()%10;// this is for making item.
while (buffer.size() > size)
{
}
buffer.push_back(num);
counter++;
return (DWORD)n;
}
这是我consumer--
DWORD WINAPI consumer(LPVOID n)
{
cout << "\nCONSUMER:The consumer is running right now" << endl;
while (buffer.empty())
{ }
int item= buffer.front();
cout << "\nCONSUMER:The consumer ate" << item << endl;
counter++;
return (DWORD)n;
}
在main-
while (counter < l)
{
hThreads[0] = CreateThread(NULL, 0, producer, (LPVOID)s, NULL, &id[0]);
hThreads[1] = CreateThread(NULL, 0, consumer, (LPVOID)l, NULL, &id[1]);
waiter = WaitForMultipleObjects(MAX_THREADS, hThreads, TRUE, INFINITE);
}
for (int i = 0; i < MAX_THREADS; i++) {
CloseHandle(hThreads[i]);
}
所以每次只产生1个。 Srand也没有工作。但它运行正确的次数。
编辑--- 所以我固定的生产和消费,以应对竞争条件:
DWORD WINAPI producer(LPVOID s)
{
WaitForSingleObject(Empty, INFINITE);
WaitForSingleObject(Mutex, INFINITE);
cout << "\nPRODUCER...." << endl;
int size = (int)s;
srand(size);
int in = rand() % 10;
cout << "THIS IS IN:::" << in << endl;
while (buffer.size() == size)
{
ReleaseMutex(Mutex);
}
buffer.push_back(in);
counter++;
cout << "\nThe producer produces " << buffer.front() << endl;
ReleaseMutex(Mutex);
ReleaseSemaphore(Full, 1, NULL);
return (DWORD)s;
}
DWORD WINAPI consumer(LPVOID l)
{
WaitForSingleObject(Full, INFINITE);
WaitForSingleObject(Mutex, INFINITE);
cout << "\nCONSUMER...." << endl;
while (buffer.empty())
{
ReleaseMutex(Mutex);
}
int out = buffer.front();
counter++;
ReleaseMutex(Mutex);
ReleaseSemaphore(Empty, 1, NULL);
return (DWORD)l;
}
但随意的事情仍保持演戏了。它只会一直重复生成一个数字(即使播种时也是如此)。
那么同步在哪里呢? – Mysticial
如果消费者没有任何东西先运行,那么生产者就会进入,然后消费者就会从中断。这取决于首先启动哪个 –
缓冲区上存在未受保护的竞争条件。这是未定义的行为。 – Mysticial