Question

我有boxIn[]类型的数组char，其中包含R R G B G B O Y O O P R个字符。 boxIn[]位于共享内存中。共享内存中还有char* p，其值为char数组中的boxIn[]之一。

程序中定义了两个信号量如下：

/* initialize semaphores */
if((sem_init(&sem, pshared, value)) == 1){      /* only 2(value) processes at the same time */
    perror("Error initializing synch semaphore\n");
    exit(1);
}
if((sem_init(&mutex, pshared, 1)) == 1){        /* only 1 processes at the same time */
    perror("Error initializing synch semaphore\n");
    exit(1);
}

我从for循环中分叉孩子：

/* fork child processes */
for(i=0; i<n ; i++){
    pid = fork();
    if(pid < 0){    /* check for error  */
        printf("Fork error.\n");
    }
    else if(pid == 0) break;    /* child processes */

}

然后我让孩子们做一些处理。目前我正在努力逐步实现目标。因此，我只更改*p的值一次。

/******************************************************/
/******************   PARENT PROCESS   ****************/
/******************************************************/
if(pid != 0){
    while(wait()>0);
}

/******************************************************/
/******************   CHILD PROCESS   *****************/
/******************************************************/
else{
    while(*p != boxIn[i]); /* wait until it's my turn */
    sem_wait(&sem);
    printf("%c boxes are being painted.\n",boxIn[i]);
    printf("Done painting.\n");
    sleep(1);
    sem_wait(&mutex);
    if(*p=='R') *p='G';
    sem_post(&mutex);
    sem_post(&sem);
    exit(1);
}

但是我确实得到了意想不到的输出：

我的期望是：

R boxes are being painted.
R boxes are being painted.
Done.
Done.
R boxes are being painted.
Done.
G boxes are being painted.
G boxes are being painted.
Done.
Done.

我得到的如下：

R boxes are being painted.
Done painting.
R boxes are being painted.
Done painting.
R boxes are being painted.
Done painting.
G boxes are being painted.
Done painting.
G boxes are being painted.
Done painting.
varaquilex@computer ~/Dropbox/Courses/BLG312E - Computer Operating Systems/hw3 $ G boxes are being painted.
Done painting.
G boxes are being painted.
Done painting.
G boxes are being painted.
Done painting.
G boxes are being painted.
Done painting.
G boxes are being painted.
G boxes are being painted.
Done painting.
Done painting.
G boxes are being painted.
Done painting.
G boxes are being painted.
Done painting.
P boxes are being painted.
Done painting.
P boxes are being painted.
Done painting.
P boxes are being painted.
Done painting.
P boxes are being painted.
Done painting.
P boxes are being painted.
Done painting.
^C

问题是：为什么即使G数组中只有2个G框，我也会获得超过2个boxIn[]框，更重要的是如何我甚至画了P个框，更不用说数组中只有一个P框了，我画了很多P个框？

注意：我已经尝试了很长时间。更不用说我使用shift + C在此之前停止了许多程序。我重新启动终端并获得此输出。更有意思的是，我试图打印i的值以及正在绘制的框，我看到输出混合了！我的意思是，即使G boxes(3) are being painted为G boxes are being painted，也有一些行显示printf()而其他一些行printf("%c boxes(%d) are being painted",boxIn[i],i);。这很尴尬，我重新启动了操作系统并从同一程序获得了不同的输出。有人可以解释为什么会发生这种情况，我该如何防止这种情况再次发生？

Answer 1

为了使用进程共享信号量，它们必须位于共享内存中。你的不是。所以在fork之后，每个进程都有自己的信号量副本。分配共享内存的最简单方法是：

sem_t *sem = mmap(0, sizeof(sem_t), PROT_READ|PROT_WRITE,
                  MAP_SHARED|MAP_ANONYMOUS, -1, 0);

然而，尽管几乎普遍可用，但POSIX中没有指定匿名映射，因此，正式地说，“不可移植”。另一种可移植的方法是创建临时文件或POSIX共享内存对象。

或（这是推荐的方式），使用sem_open代替sem_init。

分叉进程，信号量，为什么这个输出？

1 个答案: