Question

我的问题是：

1。）如何让父进程永远死？我知道这没有完成，因为父亲是第一个被运行的pid，但我不知道如何改变它。

2.。）如何让我的子进程像他一样同时执行？我甚至把这个数字提高到很高，看它是不是巧合，但似乎不是。

编辑：解决方案

1。）在内部默认
中添加了两次wait（NULL） 2.）正在发生。用睡眠（1）来证明它。

我的代码如下

#include <stdio.h>
int main() {
    int pid, i;
    pid = fork();
    switch(pid) {
        case -1:
            // error
            printf("Fork error");
            break;
        case 0:
            // child process
            printf("First child is born, my pid is %d\n", getpid());
            for(i=1; i<10; i++)
                printf("First child executes iteration %d\n", i);
            printf("First child dies quietly.\n");
            break;
        default:
            // parent process
            printf("Parent process is born, my pid is %d\n", getpid());
            pid = fork();
            switch(pid) {
                case -1:
                    // error
                    printf("Fork error");
                    break;
                case 0:
                    // child process
                    printf("Second child is born, my pid is %d\n", getpid());
                    for(i=1; i<10; i++)
                        printf("Second child executes iteration %d\n", i);
                    printf("Second child dies quietly.\n");
                    break;
                default:
                    // parent process
                    printf("Parent process dies quietly.");
        }
    }
    return 0;
}

我的输出总是如下：

Parent process is born, my pid is 7847  
First child is born, my pid is 7848  
First child executes iteration: 1  
First child executes iteration: 2  
First child executes iteration: 3  
First child executes iteration: 4  
First child executes iteration: 5  
First child executes iteration: 6  
First child executes iteration: 7  
First child executes iteration: 8  
First child executes iteration: 9  
First child executes iteration: 10  
First child dies quietly.  
Parent process dies quietly.  
Second child is born, my pid is 7849  
Second child executes iteration 1  
Second child executes iteration 2  
Second child executes iteration 3  
Second child executes iteration 4  
Second child executes iteration 5  
Second child executes iteration 6  
Second child executes iteration 7  
Second child executes iteration 8  
Second child executes iteration 9  
Second child executes iteration 10  
Second child dies quietly.

我的任务是：

编写一个C程序（“procs.c”），它创建三个进程：一个创建两个子进程的父进程。

第一个孩子应该做到以下几点：

显示“第一个孩子出生，我的pid是......”
显示十次“第一个孩子执行迭代X”的消息，其中X是迭代次数
显示“第一个孩子安静地死去。”

第二个孩子应该做到以下几点：

显示“第二个孩子出生，我的pid是......”
显示十次“第二个孩子执行迭代X”的消息，其中X是迭代次数
显示“第二个孩子安静地死去。”

父进程应执行以下操作：

显示“父进程诞生，我的pid是......”
创建第一个孩子
创建第二个孩子
显示“父进程悄然死亡。”

使用gcc编译程序并命名可执行文件“procs”。多次执行该程序，并注意两个孩子的输出如何交织。

该程序的可能输出是：

nova> ./procs

Parent process is born, my pid is 7847  
First child is born, my pid is 7848  
First child executes iteration: 1  
First child executes iteration: 2  
First child executes iteration: 3  
First child executes iteration: 4  
First child executes iteration: 5  
Second child is born, my pid is 7849  
Second child executes iteration 1  
Second child executes iteration 2  
Second child executes iteration 3  
First child executes iteration: 6  
Second child executes iteration 4  
Second child executes iteration 5  
Second child executes iteration 6  
First child executes iteration: 7  
Second child executes iteration 7  
Second child executes iteration 8  
Second child executes iteration 9  
Second child executes iteration 10  
Second child dies quietly.  
First child executes iteration: 8  
First child executes iteration: 9  
First child executes iteration: 10  
First child dies quietly.  
Parent process dies quietly.

Answer 1

在退出之前，相应的父进程应等待（使用wait() or waitpid()或特定于平台的变体）让其子进程死亡。
并发执行需要调度程序有机会运行。您可以通过适当的睡眠（微睡眠，nano-sleep）操作来强制解决问题。一些系统调用会有所帮助（因此fflush(0)可能有所帮助）。

#include <stdio.h>
#include <sys/wait.h>
#include <time.h>
#include <unistd.h>

int main(void)
{
    int pid, i;
    struct timespec tw = { .tv_sec = 0, .tv_nsec = 10000000 };
    pid = fork();
    switch(pid)
    {
        case -1:
            printf("Fork error");
            break;
        case 0:
            printf("First child is born, my pid is %d\n", getpid());
            for(i=1; i<10; i++)
            {
                printf("First child executes iteration %d\n", i);
                nanosleep(&tw, 0);
            }
            printf("First child dies quietly.\n");
            break;
        default:
            printf("Parent process is born, my pid is %d\n", getpid());
            pid = fork();
            switch(pid)
            {
                case -1:
                    printf("Fork error");
                    break;
                case 0:
                    printf("Second child is born, my pid is %d\n", getpid());
                    for(i=1; i<10; i++)
                    {
                        printf("Second child executes iteration %d\n", i);
                        nanosleep(&tw, 0);
                    }
                    printf("Second child dies quietly.\n");
                    break;
                default:
                    printf("Parent process waiting for children.\n");
                    int corpse;
                    int status;
                    while ((corpse = waitpid(0, &status, 0)) > 0)
                        printf("Child %d died with exit status 0x%.4X\n", corpse, status);
                    printf("Parent process dies quietly.\n");
                    break;
            }
    }
    return 0;
}

示例输出：

Parent process is born, my pid is 46624
First child is born, my pid is 46625
First child executes iteration 1
Parent process waiting for children.
Second child is born, my pid is 46626
Second child executes iteration 1
First child executes iteration 2
Second child executes iteration 2
First child executes iteration 3
Second child executes iteration 3
First child executes iteration 4
Second child executes iteration 4
Second child executes iteration 5
First child executes iteration 5
Second child executes iteration 6
First child executes iteration 6
Second child executes iteration 7
First child executes iteration 7
Second child executes iteration 8
First child executes iteration 8
Second child executes iteration 9
First child executes iteration 9
First child dies quietly.
Second child dies quietly.
Child 46625 died with exit status 0x0000
Child 46626 died with exit status 0x0000
Parent process dies quietly.

请注意，10毫秒延迟几乎强制交替执行。如果没有类似的东西，你会遇到系统及其调度程序的特性，很多现代机器太快了！

Answer 2

fork()创建一个新的子进程，该进程独立于父进程运行。

如果您希望父进程等待子进程完成，您可以使用系统调用wait，传递一个PID，它将阻塞直到进程完成。

请参阅文章wait(System Call) on Wikipedia

为什么我的进程并不同时运行？

2 个答案: