Question

#include <stdio.h>
#include <fcntl.h>
#include <string.h>
#include <unistd.h>
#include <sys/ipc.h>
#include <sys/shm.h>
#include <sys/sem.h>

int main()
{
  key_t key, k, shmid;
  int ret, semid, i = 0;
  struct sembuf op[1];
  void *shmptr = (void *) 0;

  union semun {
    int val;
    struct senid_ds *buf;
    unsigned short *array;
  };

  union semun arg;

  /* Creating a shared memory */
  k = ftok(".", 1);

  if (-1 == k) {
    perror("\n ftok \n");
    return 1;
  }

  shmid = shmget(k, 1000, IPC_CREAT | 0744);

  if (-1 == shmid) {
    perror("\n Shared memory creation failed \n");
    return 2;
  }

  shmptr = shmat(shmid, (void *) 0, 0);

  if (NULL == shmptr) {
    perror("\n Process cant attach to shared memory segment! \n");
    return 3;
  }

  /* Creating semaphore */
  key = ftok(".", 4);

  if (-1 == key) {
    perror("\n ftok \n");
    return 4;
  }

#if 0
  semid = semget(key, 1, IPC_CREAT);
#else
  semid = semget(key, 1, 0);
#endif

  if (-1 == semid) {
    perror("\n Cannot create semaphore \n");
    return 5;
  }

#if 0
  arg.val = 1;
  ret = semctl(semid, 0, SETVAL, arg);
  printf("\n ret : %d \n", ret);
#endif

  op[0].sem_num = 0;
  op[0].sem_op = -1;
  op[0].sem_flg = SEM_UNDO;

  ret = semop(semid, op, 1);

  // printf("\n ret : %d semop : %d\n",ret,op[0].sem);
  printf("\n Process 1 locked sem1 \n");

  /* Do Operation on shared resource */
  while (i < 4) {
    strcpy((char*) shmptr, "BDC");
    printf("\n %s \n", (char *) shmptr);
    sleep(2);
    printf("\n After sleep \n");
    i++;
    shmptr++;
  }

  op[0].sem_op = 1;
  ret = semop(semid, op, 1);
  printf("\n Process A unlocked sem1 \n");

  return 0;
}

我尝试了一个示例程序来锁定共享内存段。我尝试了同一个程序的2个实例。查看信号量是否正确锁定。但是，当process1已经锁定时，上述程序总是允许其他进程使用共享内存。无法确定错误的位置。

Answer 1

您应该启用IPC_CREAT语句和按位 - 或者在第3个参数中启用必要但缺少的权限（至少S_IRWXU）。

您应该启用SETVAL语句将信号量初始化为1，但只根据外部输入执行该语句（即，使用一个参数调用该程序，使其创建信号量并初始化它）;否则，程序应该假定信号量已经存在。

您需要检查semop()来电的返回值。

问题在共享内存中锁定信号量

1 个答案: