while循环中fork()中的fork()

Question

请考虑在这里查看我的代码。我的计划是我有一个while循环。在while循环中，我执行了一条for语句。在for循环之后，我使用fork。现在我有了一个家长和一个child1。在parent内部，我执行了另一个fork，给了我一个parent和child2。现在我的问题是：

1)为什么在成功时，fork1 x=3会打印两次？

2)对于x=2，x=3也出现了同样的问题。它说fork1和fork2是成功的，但没有同时进入child1和child2。它跳过了n=waitpid(-1, &status, 0);行，继续打印n，然后是x--并转到x=1;

3)对于x=1，我认为输出真的搞混了，比如为什么在发送1和发送2之间打印"child1 pid=4783“。fork 1也打印了两次。

请帮我解决这些问题。我一直在看帖子，但我似乎看不到类似的问题。我还能错过什么呢？非常感谢!下面是我的代码片段：

while(x>0)
{
    printf("x=%d\n", x);
    for(i=0; i<3; i++)
    {
        printf("SENDING %d\n", i);
    }

    pid1=fork();
    printf("fork1 successful\n");
    if(pid1>0)
    {
        printf("RECEIVING %d\n", i);
        pid2=fork();
        if(pid2>0)
        {
            printf("fork2 successful\n");
            n=waitpid(-1, &status, 0);
            printf("%d\n", n);

            if(n==pid1) //sleep done
            {
                kill(pid2, SIGKILL);
                printf("Child1 ran. Child2 killed.\n\n");
            }

            else if(n==pid2) //scanf received               
            {
                kill(pid1, SIGKILL);
                printf("Child2 ran. Child1 killed.\n\n");
            }
        }    
        else
        {
            printf("child2 pid=%d\n", getpid());            
            scanf("%d", &y);
            exit(1);
        }
    }
    else
    {
        printf("child1 pid=%d\n", getpid());
        sleep(5);
        exit(0);
    }
x--;
}

以及以下结果：

x=3

SENDING 0

SENDING 1

SENDING 2

fork1 successful

RECEIVING 3

fork1 successful

child1 pid=4781

fork2 successful

child2 pid=4782

4781

Child1 ran. Child2 killed.

x=2

SENDING 0

SENDING 1

SENDING 2

fork1 successful

RECEIVING 3

fork1 successful

fork2 successful

4782

x=1

SENDING 0

SENDING 1

child1 pid=4783

SENDING 2

fork1 successful

RECEIVING 3

child2 pid=4784

fork2 successful

child2 pid=4786

fork1 successful

child1 pid=4785

1

4784

Answer 1

我相信您假设n=waitpid(-1, &status, 0)会暂停，直到其中一个子进程完成。在任何子进程中发生任何更改后，waitpid都将返回。如果将变量"x“添加到打印语句中，并添加一条语句来显示来自waitpid的状态返回值，则可以看到，在x=2所在的第二个循环中，waitpid语句正由前一个循环的进程的终止信号触发。在这里，事情变得更加混乱--因为进程可能会抢占彼此。在您的原始代码中，您可以看到两个针对x=1的child1进程。

x=3
[3] SENDING 0
[3] SENDING 1
[3] SENDING 2
[3] fork1 successful
[3] RECEIVING 3
[3] fork2 successful
[3] fork1 successful
[3] child1 pid=8166
[3] child2 pid=8167
[3] child1 exiting
[3] process ID 8166 returned status 0.[3] Child1 ran. Child2 killed.

x=2
[2] SENDING 0
[2] SENDING 1
[2] SENDING 2
[2] fork1 successful
[2] RECEIVING 3
[2] fork1 successful
[2] fork2 successful
[2] process ID 8167 returned status 9.x=1
[1] SENDING 0
[1] SENDING 1
[1] SENDING 2
[2] child1 pid=8171
[2] child2 pid=8172
[1] fork1 successful
[1] RECEIVING 3
[1] fork1 successful
[1] child1 pid=8173
[1] fork2 successful
[1] child2 pid=8174
[2] child1 exiting
[1] child1 exiting
[1] process ID 8171 returned status 0

解决此问题的一种方法是检查waitpid的状态

do
{
  n=waitpid(-1, &status, 0);
  printf("[%d] process ID %d returned status %d.", x, n, status);
  if (WIFEXITED(status)==0)
     printf("This is NOT an exit status, so I will keep looping....\n");
  else
     printf("\n");

 } while (WIFEXITED(status)==0);

然后我相信你会得到预期的结果：

x=3
[3] SENDING 0
[3] SENDING 1
[3] SENDING 2
[3] fork1 successful
[3] RECEIVING 3
[3] fork2 successful
[3] fork1 successful
[3] child1 pid=8267
[3] child2 pid=8268
[3] child1 exiting
[3] process ID 8267 returned status 0.
[3] Child1 ran. Child2 killed.

x=2
[2] SENDING 0
[2] SENDING 1
[2] SENDING 2
[2] fork1 successful
[2] RECEIVING 3
[2] fork2 successful
[2] process ID 8268 returned status 9.This is NOT an exit status, so I will keep looping....
[2] fork1 successful
[2] child1 pid=8270
[2] child2 pid=8271
[2] child1 exiting
[2] process ID 8270 returned status 0.
[2] Child1 ran. Child2 killed.

x=1
[1] SENDING 0
[1] SENDING 1
[1] SENDING 2
[1] fork1 successful
[1] RECEIVING 3
[1] fork1 successful
[1] child1 pid=8273
[1] fork2 successful
[1] process ID 8271 returned status 9.This is NOT an exit status, so I will keep looping....
[1] child2 pid=8274
[1] child1 exiting
[1] process ID 8273 returned status 0.
[1] Child1 ran. Child2 killed.