Question

我正在尝试做一些基本的并行处理，使用POSIX共享内存段和未命名的信号量对整数进行奇偶排序。我有几乎所有工作在这一点上，除了最后一件事：如果我在信号量锁定/解锁后不直接perror（）代码行为不同（并随后排序不正确）。如果我在信号量锁定和解锁后直接离开perror（）调用，代码会完美地对整数数组进行排序。

int semaphoreCheck = sem_init(&(sharedData->swapSem), 1, 1);

if (semaphoreCheck == -1)
{
    perror( "failed to initialize semaphore" );
    exit(EXIT_FAILURE);
}

pid_t fork1;
fork1 = fork();
if (fork1 == 0)
{
    // original.child
    pid_t fork2;
    fork2 = fork();
    if (fork2 == 0)
    {
        // child.child
        // do a portion of the sort here
        while(sharedData->evenSwap || sharedData->oddSwap)
        {
            // obtain lock on the shared vector
            // int commandCheck = shmctl(sharedID, SHM_LOCK, NULL);
            int commandCheck = sem_wait(&(sharedData->swapSem));
            perror("semaphore lock");
            // if lock was obtained
            if (commandCheck == 0)
            {
                sharedData->evenSwap = false;
                for( int index = 1; index < arraySize - 1; index +=2)
                {
                    if( sharedData->vecData[index] > sharedData->vecData[index + 1] )   
                    {
                        int temp;
                        temp = sharedData->vecData[index];
                        sharedData->vecData[index] = sharedData->vecData[index+1];
                        sharedData->vecData[index+1] = temp;
                        sharedData->evenSwap = true;
                    }       
                }
                // release lock on the shared vector
                commandCheck = sem_post(&(sharedData->swapSem));
                perror("semaphore unlock");
                if (commandCheck == -1)
                {
                    perror("failed to unlock shared semaphore");
                }
            }
            else perror("failed to lock shared semaphore");
        }
        _exit(0);
    }
    else if (fork2 > 0)
    {
        // child.parent
        // do a portion of the sort here
        while(sharedData->evenSwap || sharedData->oddSwap)
        {
            // obtain lock on the shared vector
            int commandCheck = sem_wait(&(sharedData->swapSem));
            perror("semaphore lock");
            // if lock was obtained
            if (commandCheck == 0)
            {
                sharedData->oddSwap = false;
                for( int index = 0; index < arraySize - 1; index +=2)
                {
                    if( sharedData->vecData[index] > sharedData->vecData[index + 1] )   
                    {
                        int temp;
                        temp = sharedData->vecData[index];
                        sharedData->vecData[index] = sharedData->vecData[index+1];
                        sharedData->vecData[index+1] = temp;
                        sharedData->oddSwap = true;
                    }       
                }
                // release lock on the shared vector
                commandCheck = sem_post(&(sharedData->swapSem));
                perror("semaphore unlock");
                if (commandCheck == -1)
                {
                    perror("failed to unlock shared semaphore");
                }
            }
            else perror("failed to lock shared semaphore");
        }
        _exit(0);
    }
    else
    {
        // child.error
        // forking error.
        perror("failed to fork in child");
        exit(EXIT_FAILURE);
    }
}
else if( fork1 > 0)
{
    // original.parent
    // wait for the child process to finish.
    waitpid(fork1, NULL, 0);

}
else
{
    // forking error
    perror("failed to fork");
    exit(EXIT_FAILURE);
}

我只能猜测这与信号量如果无法满足等待阻止进程有关，但我不明白perror（）调用如何修复它。

Answer 1

我认为您的问题可能与您（不）在获得信号量后检查条件仍然适用的方式有关，或者检查条件本身是错误的。

你有：

    while(sharedData->evenSwap || sharedData->oddSwap)
    {
        // obtain lock on the shared vector
        int commandCheck = sem_wait(&(sharedData->swapSem));
        perror("semaphore lock");
        // if lock was obtained
        if (commandCheck == 0)
        {
            sharedData->oddSwap = false;

获得信号量后，您应该验证sharedData->evenSwap或sharedData->oddSwap是否仍为真，如果没有，则放弃信号量。这是一个标准的习语;检查，锁定和重新检查，因为原始检查和获得锁定的时间之间的状态可能已经改变。

在此假设下，perror()调用会改变进程的时间，允许条件保持不变的时间长于perror()调用不存在的时间。所以，这里有一个时间问题，某处。

perror（）调用问题影响信号量锁定/解锁

1 个答案: