这段代码其实开头都还好,关键是末尾要结束的时候有一段代码比较费解。我想,这就是我前面说过的优先级反转问题。为了解决这一问题,在rawos版本中采取了优先级继承的方法。我们还是详细看一下逻辑本身是怎么样的,
(1)判断参数合法性;
(2)判断资源是否可取,如果可取,则在记录当前线程和优先级后返回;
(3)如果资源被自己重复申请,返回;
(4)如果线程不愿等待,返回;
(5)如果此时禁止调度,返回;
(6)如果此时优先级大于互斥量占有者的优先级,分情况处理
a)占有者处于ready的状态,那么修改它的优先级,重新加入调度队列;
b)占有者处于sleep的状态,直接修改优先级即可;
c)占有者也被pend到别的资源上面了,那么修改那个资源的pend列表,可能设计到调度顺序问题。
(7)线程把自己pend到互斥量等待队列上面;
(8)线程调用系统调度函数,切换到其他线程运行;
(9)线程再次得到运行的机会,从task获取结果后返回。
基本上上面的介绍算得上是很详细了,那么互斥量的释放基本上是一个逆操作的过程,朋友也可以思考一下应该怎么解决才好,
RAW_U16 raw_mutex_put(RAW_MUTEX *mutex_ptr) { LIST *block_list_head; RAW_SR_ALLOC(); #if (RAW_MUTEX_FUNCTION_CHECK > 0) if (mutex_ptr == 0) { return RAW_NULL_OBJECT; } #endif block_list_head = &mutex_ptr->common_block_obj.block_list; RAW_CRITICAL_ENTER(); /*Must release the mutex by self*/ if (raw_task_active != mutex_ptr->occupy) { RAW_CRITICAL_EXIT(); return RAW_MUTEX_NOT_RELEASE_BY_OCCYPY; } /*if no block task on this list just return*/ if (is_list_empty(block_list_head)) { mutex_ptr->count = 1; RAW_CRITICAL_EXIT(); return RAW_SUCCESS; } /*if priority was changed, just change it back to original priority*/ if (raw_task_active->priority != mutex_ptr->occupy_original_priority) { remove_ready_list(&raw_ready_queue, raw_task_active); raw_task_active->priority = mutex_ptr->occupy_original_priority; add_ready_list_end(&raw_ready_queue, raw_task_active); } /* there must have task blocked on this mutex object*/ mutex_ptr->occupy = list_entry(block_list_head->next, RAW_TASK_OBJ, task_list); /*the first blocked task became the occupy task*/ mutex_ptr->occupy_original_priority = mutex_ptr->occupy->priority; /*mutex resource is occupied*/ mutex_ptr->count = 0; /*Wake up the occupy task, which is the highst priority task on the list*/ raw_wake_object(mutex_ptr->occupy); RAW_CRITICAL_EXIT(); raw_sched(); return RAW_SUCCESS; } |