当尝试运行此代码时:首先它会为每个信号量打印“Process some_id
BEFORE enter”(2 次)。然后它挂起。哪里不对?
# include <sys/ipc.h>
# include <sys/sem.h>
# include <unistd.h>
# include <errno.h>
# include <stdio.h>
# include <stdlib.h>
# include <iostream>
int seminit()
{
key_t key = ftok("/bin", 1);
int semid = semget (key, 1, IPC_CREAT | IPC_EXCL | 600);
if(-1 == semid)
if(EEXIST == errno)
semid = semget(key, 1, 0);
return semid;
}
void uninit(int semid)
{
semctl(semid, 0, IPC_RMID);
}
void semlock(int semid)
{
struct sembuf p_buf;
p_buf.sem_num = 0;
p_buf.sem_op = -1;
p_buf.sem_flg = SEM_UNDO;
if(semop(semid, &p_buf, 1) == -1)
printf("semlock failed: ERRNO: %d\n", errno);
}
void semunlock(int semid)
{
struct sembuf v_buf;
v_buf.sem_num = 0;
v_buf.sem_op = 1;
v_buf.sem_flg = SEM_UNDO;
if(semop(semid, &v_buf, 1) == -1)
printf("semunlock failed: ERRNO: %d\n", errno);
}
void some_function()
{
int semid = seminit();
pid_t pid = getpid();
printf("Process %d BEFORE enter\n", pid);
semlock(semid);
printf("Process %d IN Critical section\n", pid);
sleep(10);
semunlock(semid);
printf("Process %d AFTER leave\n", pid);
uninit(semid);
}
int main(int argc, char** argv)
{
for(int i = 0; i < 2; ++i)
if(0 == fork())
some_function();
return (EXIT_SUCCESS);
}
最佳答案
似乎只生成了一个子进程(我认为这不是故意的),但我相信在主进程返回之前缺少等待,这意味着主进程将比子进程结束得更快并且让它“挂起”(这可能是问题的一部分,但可能不是漏洞问题,请检查之前的循环)。
关于c++ - 为什么 semop() 挂起?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4521607/