为什么 semop() 挂起？

Question

尝试运行此代码时：首先，它为每个信号量打印“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);
}

Answer 1

似乎只生成了一个子进程（我认为这不是故意的），但我认为在主进程返回之前缺少等待，这意味着主进程将比子进程更快地结束并且让它“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）。