c - 信号量和互斥循环

Question

我有两个函数，我使用信号量和互斥锁来解决经典的睡眠理发师问题。但是当我运行程序时，程序永远不会结束。

功能一：TA的动作（理发师举例）

void *teaching_assistant()
{       
        printf("\t<-- TA opened his office\n");
        printf("\t<-- TA waiting for student\n");
        
        while(1){
                // If there is waiting students
                if(waiting_students > 0){
                        ta_sleep_flag = 0;
                        sem_wait(&sem_stu);
                        pthread_mutex_lock(&mtx);

                        // TA giving advice to student
                        printf("\t<-- TA starts advising session\n");

                        num_chairs[next_teaching_position] = 0;
                        waiting_students--;
                        next_teaching_position = (next_teaching_position + 1) % NUM_MAX_CHAIR;

                        sleep(adviceperiod);
                        pthread_mutex_unlock(&mtx);
                        sem_post(&sem_ta);
                }
                // If there is no waiting student
                else{
                        if(ta_sleep_flag == 0){
                                printf("\t<-- TA finished advising, taking a short nap\n");
                                ta_sleep_flag = 1;
                                sleep(napperiod);
                                sem_post(&sem_ta);
                        }
                }
        }
}

功能2：学生的行为（理发师示例的客户）

void *student(void *student_id)
{
        int id_student = (int *)student_id;
        printf("std<%d> entered lab\n", id_student);

        while(1){
                // If there is a waiting student, continue waiting
                if(waiting_status(id_student) == 1){ continue; }

                // Student is programming
                printf("std<%d> programming for %d units\n", id_student, workperiod);
                sleep(workperiod);

                pthread_mutex_lock(&mtx);
                
                printf("std<%d> checking TA office\n", id_student);
                //found available cahir 
                if(waiting_students < NUM_MAX_CHAIR){
                        num_chairs[next_seating_position] = id_student;
                        waiting_students++;
                        chair++;
                        printf("std<%d> entered TA office; %d seats occupied\n", id_student, chair);
                        next_seating_position = (next_seating_position + 1) % NUM_MAX_CHAIR;

                        pthread_mutex_unlock(&mtx);

                        // Wake TA if sleeping
                        sem_post(&sem_stu);
                        sem_wait(&sem_ta);
                }
                
                // Go drink coffe if there is no available seat
                else{
                        pthread_mutex_unlock(&mtx);
                        sleep(coffeeperiod);
                        sem_post(&sem_stu);
                }
        }
}

我的预期输出是：

<--TA opened his office
<--TA waiting for a student
std <2145236736> entered lab 
std <2145236736>programming for 10 units
std <2145236736> checking TA office
std <2145236736> entered TA office; 1 seats occupied
<--TA starts advising session
std <2145236736> asking questions
std <2145236736> leaving TA office
std <2145236736> signing off; active students= 0
<--TA finished advising, taking a short nap
<--TA signing off

但是当我编译代码时，我得到了无限循环。

标签： cmultithreadingoperating-systemmutexsemaphore