c - CreateSemaphore() API 中的 lmaxcount 参数有什么用？

Question

我正在学习 Windows 操作系统，并且正在编写标准的消费者生产者问题。我有一个用于资源计数的信号量和一个用于同步的互斥锁。我已经在 CreateSemaphore() 中传递了最大计数值 50，因此它不应允许生产者创建超过 50 个资源。但是当我运行代码时，它远远不止于此。我是否错误地理解了 max count 参数的使用？我也贴代码。请帮我解决这个问题。

#include<stdio.h>
#include<windows.h>

DWORD WINAPI consumerThread(LPVOID args);
DWORD WINAPI producerThread(LPVOID args);

int shared;
HANDLE hMutex;
HANDLE hSemaphore;
HANDLE hConsumer;
HANDLE hProducer;
DWORD dwConsumerId,dwProducerId;

#define MAX_COUNT 50
#define MIN_COUNT 0
int main()
{
    if(!(hMutex=CreateMutex(NULL,0,NULL)))
    {
        puts("Error:: unable to create Mutex!!");
        ExitProcess(GetLastError());
    }
    if(!(hSemaphore=CreateSemaphore(NULL,MIN_COUNT,MAX_COUNT,NULL)))
    {
        puts("Error:: unable to create Semaphore object!!");
        ExitProcess(GetLastError());
    }
    if(!(hConsumer=CreateThread(NULL,0,consumerThread,NULL,0,&dwConsumerId)))
    {
        puts("Error:: unable to create consumer Thread!!");
        ExitProcess(GetLastError());
    }
    if(!(hProducer=CreateThread(NULL,0,producerThread,NULL,0,&dwProducerId)))
    {
        puts("Error:: unable to create producer Thread!!");
        ExitProcess(GetLastError());
    }
    WaitForSingleObject(hConsumer,INFINITE);
    WaitForSingleObject(hProducer,INFINITE);
    CloseHandle(hMutex);
    CloseHandle(hSemaphore);
    CloseHandle(hConsumer);
    CloseHandle(hProducer);
    return 0;
}

DWORD WINAPI consumerThread(LPVOID args)
{

    while(1)
    {
        WaitForSingleObject(hSemaphore,INFINITE);
        WaitForSingleObject(hMutex,INFINITE);
        shared--;
        printf("Consumer  = %d\n",shared);
        ReleaseMutex(hMutex);
        //Sleep(1000);
    }
}

DWORD WINAPI producerThread(LPVOID args)
{
    if(!SetThreadPriority(hProducer,THREAD_PRIORITY_HIGHEST))
    {
        printf("Error:: Unable to set the thread priority level!!\n");
        ExitProcess(GetLastError());
    }
    while(1)
    {
        WaitForSingleObject(hMutex,INFINITE);
        shared++;
        printf("Producer =%d\n",shared);
        ReleaseMutex(hMutex);
            ReleaseSemaphore(hSemaphore,1,NULL);


    }
}

Answer 1

@Hans Passant 和@Igor Tandetnik 在评论中回答了这个问题。要点是，如果您尝试将信号量的值增加到超过 lmaxcount 参数中指定的值，ReleaseSemaphore() API 将失败，而不是阻塞调用线程。所以你应该检查它的返回值，因为我没有这样做并陷入麻烦。 :)