perror() 调用影响信号量锁定/解锁的问题

Question

我正在尝试使用 POSIX 共享内存段和未命名信号量进行一些基本的并行处理，以对整数进行奇偶排序。此时我几乎已经完成了所有工作，除了最后一件事：如果我在信号量锁定/解锁之后不直接 perror() ，则代码的行为会有所不同（并且随后排序不正确）。如果我在信号量锁定和解锁之后直接保留 perror() 调用，则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);
}

我只能猜测这与如果无法完成等待则信号量如何阻止进程有关，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在获得信号量后（不）检查条件是否仍然适用的方式有关，或者检查条件本身是错误的。

您有：

    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，您可能应该验证 sharedData->evenSwap 或 sharedData->oddSwap 是否仍然为 true，如果不是，则放弃信号量。这是一个标准的习语；您检查、锁定并重新检查，因为在原始检查和获得锁定之间状态可能已发生变化。

在此假设下，perror() 调用会改变进程的计时，从而使条件保持不变的时间比不存在 perror() 调用时的时间更长。所以，这里某个地方存在时间问题。