如何同时派生父进程的最多5个子进程？

Question

我有下面的代码，我试图一次只允许最多5个孩子运行，但我不知道当孩子退出时如何减少孩子的数量。

struct {
   char *s1;
   char *s2;
} s[] = {
  {"one", "oneB"},
  {"two", "twoB"},
  {"three", "thr4eeB"},
  {"asdf", "3th43reeB"},
  {"asdfasdf", "thr33eeB"},
  {"asdfasdfasdf", "thdfdreeB"},
  {"af3c3", "thrasdfeeB"},
  {"fec33", "threfdeB"},
  {NULL, NULL}
};

int main(int argc, char *argv[])
{
int i, im5, children = 0;
int pid = fork();
for (i = 0; s[i].s2; i++)
{
    im5 = 0;
    switch (pid)
    {
        case -1:
        {
            printf("Error\n");
            exit(255);
        }
       case 0:
       {
            printf("%s -> %s\n", s[i].s1, s[i].s2);
            if (i==5) im5 = 1;
            printf("%d\n", im5);
            sleep(i);
            exit(1);
        }
        default:
        {   // Here is where I need to sleep the parent until chilren < 5 
            // so where do i decrement children so that it gets modified in the parent process?
            while(children > 5)
                sleep(1);
            children++;
            pid = fork();
        }
    }
}
return 1;
}

修订的版本似乎是基于评论而工作的

struct {
   char *s1;
   char *s2;
} s[] = {
  {"one", "oneB"},
  {"two", "twoB"},
  {"three", "thr4eeB"},
  {"asdf", "3th43reeB"},
  {"asdfasdf", "thr33eeB"},
  {"asdfasdfasdf", "thdfdreeB"},
  {"af3c3", "thrasdfeeB"},
  {"fec33", "threfdeB"},
  {NULL, NULL}
};

pthread_mutex_t children_count_lock;
int children = 0;

int main(int argc, char *argv[])
{
int i, im5;
int pid = fork();
for (i = 0; s[i].s2; i++)
{
    im5 = 0;
    switch (pid)
    {
        case -1:
        {
            printf("Error\n");
            exit(255);
        }
       case 0:
       {
            printf("%s -> %s\n", s[i].s1, s[i].s2);
            if (i==5) im5 = 1;
            printf("%d\n", im5);
            sleep(i);

            pthread_mutex_lock(&children_count_lock);
            children = children - 1;
            if (children < 0) children = 0;
            pthread_mutex_unlock(&children_count_lock);

            exit(1);
        }
        default:
        {   
            if (children > 4)
                wait();

            pthread_mutex_lock(&children_count_lock);
            children++;
            pthread_mutex_unlock(&children_count_lock);

            pid = fork();
        }
    }
}
return 1;
}

Answer 1

wait()系列函数将挂起父进程，直到子进程退出(这样做而不是休眠)。

不，你根本不需要临界区--子节点和父节点不共享内存。在默认情况下，您所需要的全部内容如下所示：

    default:
    {   
        children++;  // Last fork() was successful

        while (children >= 5)
        {
            int status;
            // Wait for one child to exit
            if (wait(&status) == 0)
            {
                children--;
            }
        }

        pid = fork();
    }

(忘记我之前说的关于将children初始化为1的内容，我没有注意到children++应该在while循环之前)。

Answer 2

一旦您运行了最初的5个子进程，您可能希望使用wait()来等待子进程完成，然后您就可以启动一个新的子进程。

Answer 3

在父级中调用wait()。这将一直阻塞，直到其中一个子进程退出。