使用信号量时等待分叉进程

I'm trying to code an exercise in C Linux where I have one semaphore with 2 spots and "n" processes entered by argument. I need that the first 2 processes use the semaphore using the 2 spots for 5 secs each and then leave the semaphore for the other remaining processes to do their stuff. The problem is that not all the other processes wait for the semaphore to be free and some of them show a semaphore error (look at the results at the bottom). I believe the problem is on the waits for the child processes, because I have a waitipid and a wait function, but I need that if there's a free spot in the semaphore, any child process running could use it. Here's the code:

我正在尝试在C Linux中编写一个练习，其中我有一个信号量，有2个点，并且“n”进程通过参数输入。我需要前两个进程使用信号量使用2个点，每个5秒，然后留下信号量用于其他剩余的进程来完成它们的工作。问题是并非所有其他进程都等待信号量空闲，其中一些显示信号量错误（查看底部的结果）。我认为问题在于等待子进程，因为我有waitipid和wait函数，但我需要如果信号量中有一个空闲点，任何运行的子进程都可以使用它。这是代码：

//gcc SemaphoreExample.c -o s
//./s 5

#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <string.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <unistd.h>
#include <sys/wait.h>
#include <sys/ipc.h>
#include <sys/sem.h>

void semaphoreTask(int semid, int semnum, int semBusy)
{
    struct sembuf data;
    data.sem_num = semnum;
    data.sem_flg = 0;
    data.sem_op = semBusy;

    if(semop(semid,&data,1) == -1)
    {
        printf("\nSemaphore Error\n");
        exit(-1);
    }
}

int main(int argc, char *argv[])
{
    int i, fdSemaphore, quantity, fdsemctl, j;
    pid_t pid[15];

    system("clear");

    if(argc-1 < 1)
    {
        printf("Some arguments are missing\n");
        return EXIT_FAILURE;
    }

    printf("Number of arguments entered:\n\nargc: %d\n\nValues from the arguments:\n\n",argc-1);

    for(i=0;i<argc;i++)
    {
        printf("argv[%d]: %s\n",i,argv[i]);
    }

    printf("\n\n");

    fdSemaphore = semget(1234,1,IPC_CREAT|0777);

    if(fdSemaphore == -1)
    {
        printf("\nError creating the Semaphore\n");
        return EXIT_FAILURE;
    }

    fdsemctl = semctl(fdSemaphore,0,SETVAL,2);

    if(fdsemctl == -1)
    {
        printf("\nError opening the Semaphore\n");
        return EXIT_FAILURE;
    }

    quantity = atoi(argv[1]);

    for(i=0;i<quantity;i++)
    {
        pid[i] = fork();

        if(pid[i] == -1)
        {
            printf("\nError creating the Child Process\n");
            return EXIT_FAILURE;
        }

        if(pid[i] == 0)
        {
            semaphoreTask(fdSemaphore,0,-1);
            printf("\n[%d] I go to sleep\n",getpid());
            sleep(5);
            printf("\n[%d] I wake up\n",getpid());
            semaphoreTask(fdSemaphore,0,1);
        }
        else
        {
            //printf("\nJust wait\n");
            waitpid(pid[i],NULL,WNOHANG);
        }
    }

    for(j=0;j<quantity;j++)
    {
        wait(NULL);
    }

    semctl(fdSemaphore,0,IPC_RMID);

    return EXIT_SUCCESS;
}

This is the result I got:

这是我得到的结果：

Result:

Number of arguments entered:

argc: 1

Values from the arguments:

argv[0]: ./s
argv[1]: 5


[2845] I go to sleep

[2844] I go to sleep

[2845] I wake up

[2844] I wake up

Semaphore Error

[2843] I go to sleep

Semaphore Error

Semaphore Error

[2843] I wake up

Semaphore Error

Should I use wait or waitpid only?

我应该只使用wait或waitpid吗？

2 个解决方案

#1

The problem is semaphores are getting removed by forked children.

问题是分叉儿童正在消除信号量。

After line

排队后

semaphoreTask(fdSemaphore,0,1);

add

加

exit(0);

#2

It is much simpler to implement what you want if you use sem_post and sem_wait calls. I am on a OpenBSD system and I am assuming that Linux has the same thing.

如果使用sem_post和sem_wait调用，实现所需的内容要简单得多。我在OpenBSD系统上，我假设Linux有相同的东西。

#1