C-共享内存和信号量

Question

我想创建一个共享内存和信号量的C程序。应该有两个子进程。两个孩子都有一个不同的int数。然后有一个目标号码，应该写在共享内存中。现在两个孩子都应该从进球数中减去他们的数字，直到进球数低于或等于0。我不希望出现比赛条件。这就是为什么我尝试使用信号量。但对我没用。下面是我的代码：

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

#define SEG_SIZE sizeof(int)
#define NUM_OF_CHILDS 2

int main(int argc, char *argv[]){

    int i, shm_id, sem_id, *shar_mem;
    int pid[NUM_OF_CHILDS];
    long waittime = 100;
    unsigned short marker[1];

    /* Define the numbers and the goal number */

    int numbers[2] = {28, 23};
    int goal = (numbers[0] + numbers[1]) * 4;   

    /* Create semaphor */

    if((sem_id = semget(IPC_PRIVATE, 1, IPC_CREAT|0644)) == -1){

        perror("semget()");
        exit(EXIT_FAILURE);

    }

    marker[0] = 1;

    /* All sem's to 1 */

    semctl(sem_id, 1, SETALL, marker);

    /* Create shared memory */

    if((shm_id = shmget(IPC_PRIVATE, SEG_SIZE, IPC_CREAT|0600)) == -1){

        perror("shmget()");
            exit(EXIT_FAILURE); 

    }
    if((shar_mem = (int *)shmat(shm_id, 0, 0)) == (int *) -1){

        perror("shmat()");
        exit(EXIT_FAILURE); 

    }
    *shar_mem = goal;

    /* Create child processes */

    for(i = 0; i < NUM_OF_CHILDS; i++){

        pid[i] = fork();
        if(pid[i] < 0){

            printf("Error!\n");
            exit(1);

        }
        if(pid[i] == 0){
            int count = 0;  
            /* Child processes */

            /* Structs for semaphor */

            struct sembuf enter, leave;

            enter.sem_num = leave.sem_num = 0;      
            enter.sem_flg = leave.sem_flg = SEM_UNDO;
            enter.sem_op = -1;              /* DOWN-Operation */
            leave.sem_op = 1;               /* UP-Operation */

            /* Join critical area */

            semop(sem_id, &enter, 1);

            while(*shar_mem > 0){

                usleep(waittime);
                *shar_mem -= numbers[i];

                count++;
            }

            printf("%i\n", count);

            /* Leave critical area */

            semop(sem_id, &leave, 1);

            exit(0);

        }

    }

    /* Wait for childs. */

    for(i = 0; i < NUM_OF_CHILDS; i++){

        waitpid(pid[i], NULL, 0);

    }

    /* Is goal equal 0 or lower? */

    int returnv;

    if(*shar_mem == 0){

        /* No race conditions */

        returnv = 0;

    }
    else {

        /* Race conditions */

        returnv = 1;

    }

    /* Close shared memory and semaphores */

    shmdt(shar_mem);
    shmctl(shm_id, IPC_RMID, 0);
    semctl(sem_id, 0, IPC_RMID);

    return returnv;

}

Answer 1

将while循环调整为如下所示，以便在减去一次后离开临界部分：

for ( ; ; ) {
  usleep(waittime);
  semop(sem_id, &enter, 1);
  if (*shar_mem <= 0) {
    semop(sem_id, &leave, 1);
    break;
  }
  *shar_mem -= numbers[i];
  semop(sem_id, &leave, 1);
  count++;
}

但正如我在评论中所说的，不能保证两个孩子交替减去他们的数字，也就是说，结果可能小于零