有没有办法改变用fork()创建的子进程的执行顺序? [英] Is there a good way to alter execution order of child processes created with fork()?

问题描述

我正在寻找创建可以控制其处理顺序的子进程.

I am looking for creating child processes for which I can control their order of processing.

简单的例子:

• 父母用叉子创造了两个孩子
• 儿童

1. 第一个孩子打印消息2"
2. 第二个孩子打印消息1"

完成后，父级会打印结束"

由于我们无法确定将首先执行哪个进程，因此最终结果很有可能是:

Because of the fact that we can't know for sure which process will be executed first, there are high chances that the final result would be:

消息2

消息1

结束

我正在尝试确保第二个孩子在第一个孩子之前执行打印，而父母在所有孩子之后执行打印.

I am trying to make sure that the second child executes the print before the first child and that the parent executes its print after all the children.

对于父母来说，使用wait()/waitpid()函数非常容易.但是，带孩子们似乎更困难.

For the parent it's quite easy with the wait()/waitpid() functions. However it seems harder with the children.

这是我为实现目标而执行的想法:

Here is an implementation of my ideas to achieve the objective:

(注意:我对创建子进程还很陌生，在此实现中我可能会误解了事情)

(note: I'm still quite new to the creation of child processes and I may have misunderstood things in this implementation)

#include <stdlib.h>
#include <stdio.h>

#include <sys/types.h>
#include <signal.h>
#include <unistd.h>

static int init = 0;

void setInitFinished(int sig)
{
    if (sig == SIGUSR1)
        init = 1;
}

int main()
{
    signal(SIGUSR1, setInitFinished);
    
    pid_t pid1, pid2;
    int status1, status2;
    
    // CHILD 1
    if (!(pid1 = fork()))
    {
        while (!init); // Waiting all children to be initiated

        // Once all children created, we wait for child 2 to print its message
        int pidOfChild2 = getpid()+1; // I checked, the PID is correct
        waitpid(pidOfChild2, &status1, 0);
        
        printf("MESSAGE 2\n");
        exit(0);
    }
    
    // CHILD 2
    if (!(pid2 = fork()))
    {
        while (!init); // Waiting all children to be initiated
        
        // No need to wait since it's the first message to be printed
        printf("MESSAGE 1\n");
        exit(0);
    }
    
    // PARENT

    // All children have been created, tell it to all the children
    kill(pid2,SIGUSR1);
    kill(pid1,SIGUSR1);
    
    // When every child has finished its work, continue parent process
    waitpid(pid1, &status1, 0);
    waitpid(pid2, &status2, 0);
    
    printf("Parent end\n");
    
    return 0;
}

在孩子1中，我试图通过 waitpid(pidOfChild2，...)等待孩子2；但这似乎不起作用.

In the child 1 I am trying to wait for the Child 2 with waitpid(pidOfChild2, ...); but it doesn't seem to work.

我仍然在发现fork的功能，所以我很确定会误会这里的很多事情.

I'm still discovering the fork functionalities so I'm pretty sure to misunderstand a lot of things here.

注意:我想避免使用sleep()，它可以工作，但效果不佳

NB: I want to avoid using sleep(), it could work but it's not pretty

推荐答案

您需要使用实际的进程间通信来实现此目的.

You need to use actual inter-process communication to achieve this.

您似乎认为waitpid()函数与等待进程打印输出有关，但这根本不起作用.

You seem to think that the waitpid() function has something to do with waiting for a process to print output, but that's not at all what it does.

在父级中创建信号量，并将其传递给两个孩子，并让他们一个孩子在打印之前等待信号量，另一个孩子在完成打印后向信号量发送消息.

Create a semaphore in the parent, pass it to both children, and have one child wait on the semaphore before printing and the other one messaging the semaphore after it's done printing.

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