[RTT例程练习] 1.7 优先级翻转之优先级继承

RTT 的mutex 实现了优先级继承算法，可用其来解决优先级反转的问题。

还是来自官网：

thread2和worker线程虽然优先级比thread1要高，但是这两个线程均在进程开始出就执行了延时函数，于是轮到 thread1 执行，然后 thread1获得互斥量，thread2延时结束后，虽然它的优先级高于thread1，但是它所需的互斥量被thread1占有了，它无法获得所需的互斥量以便继续运行。在此时，系统的优先级继承算法也会起作用，将thread1的优先级提升到与thread2一致，验证方法是在thread1 release互斥量之前插入tid2->currentpriority 是否等于 tid1->currentpriority的判断语句，当然此时的结果是相等的。当 thread1 优先级被提升到和 thread2 一样后，worker 线程优先级因为低于 thread1 的优先级而不再能够抢占 thread1, 从而保证避免优先级反转现象发生。
所以说，优先级反转的问题可以通过优先级继承来解决，在RT-Thread 的 mutex 中实现了优先级继承算法。

程序：

#include <rtthread.h>

static rt_mutex_t mutex = RT_NULL;
static rt_uint8_t t1_count, t2_count, worker_count;
static rt_thread_t t1, t2, worker;

static void thread1_entry(void *parameter)
{
    rt_err_t result;
    
    result = rt_mutex_take(mutex, RT_WAITING_FOREVER);
    rt_kprintf("thread1: got mutex \n");
    
    if (result != RT_EOK)
        return;
    
    for (t1_count = 0; t1_count < 5; t1_count++)
    {
        rt_kprintf("thread1: count: %d\n", t1_count);
    }
    //if (t2->currentpriority != t1->currentpriority)
    {
        rt_kprintf("thread1: released mutex \n");
        rt_mutex_release(mutex);
    }
}

static void thread2_entry(void *parameter)
{
    rt_err_t result;
    
    rt_thread_delay(5);
    
    result = rt_mutex_take(mutex, RT_WAITING_FOREVER);
    rt_kprintf("thread2: got mutex\n ");
    for (t2_count=0; t2_count<5; t2_count++)
    {
        rt_kprintf("thread2: count: %d\n", t2_count);
    }
}

static void worker_thread_entry(void *parameter)
{
    rt_thread_delay(5);
    
    for (worker_count = 0; worker_count < 5; worker_count++)
    {
        rt_kprintf("worker: count: %d\n", worker_count);
        rt_thread_delay(5);
    }
}

int rt_application_init()
{
    mutex = rt_mutex_create("mutex", RT_IPC_FLAG_FIFO);
    if (mutex == RT_NULL)
    {
        return 0;
    }
    
    t1_count = t2_count = 0; 
    
    t1 = rt_thread_create("t1",
        thread1_entry, RT_NULL,
        512, 7, 10);
    if (t1 != RT_NULL)
        rt_thread_startup(t1);
        
    t2 = rt_thread_create("t2", 
        thread2_entry, RT_NULL,
        512, 5, 10);
    if (t2 != RT_NULL)
        rt_thread_startup(t2);
        
    worker = rt_thread_create("worker",
        worker_thread_entry, RT_NULL,
        512, 6, 10);
    if (worker != RT_NULL)
        rt_thread_startup(worker);
        
        
    return 0;
}
/*@}*/  

输出结果：

thread1: got mutex
thread1:count: 0
thread1:count: 1
thread1:count: 2
thread1:count: 3
thread1:count: 4
thread1: released mutex
thread2: got mutex
thread2: count: 0
thread2: count: 1
thread2: count: 2
thread2: count: 3
thread2: count: 4
worker:count: 0
worker:count: 1
worker:count: 2
worker:count: 3
worker:count: 4

thread1 再取得互斥信号量之后，其优先级就变为比thread2 高，便不会发生优先级反转的现象了。