[RTT例程练习] 1.4 线程优先级抢占

RTT 是抢占式的RTOS,高优先级的线程会先执行。

这个例程显示了是如何抢占的。

解释我懒得写了,下面这段来自官网论坛:

因为更高的优先级,thread1率先得到执行,随后它调用延时,时间为3个系统tick,于是thread2得到执行。可以从打印结果中发现一个规律,在第一次thread2了打印两次thread1会打印一次之后,接下来的话thread2每打印三次thread1会打印一次。对两个线程的入口程序进行分析可以发现,在thread1 3个系统tick的延时里,thread2实际会得到三次执行机会,但显然在thread1的第一个延时内thread2第三次执行并没有执行结束,在第三次延时结束以后,thread2本应该执行第三次打印计数的,但是由于thread1此时的延时也结束了,而其优先级相比thread2要高,所以抢占了thread2的执行而开始执行。当thread1再次进入延时时,之前被抢占的thread2的打印得以继续,然后在经过两次1个系统tick延时和两次打印计数后,在第三次系统tick结束后又遇到了thread1的延时结束,thread1再次抢占获得执行,所以在这次thread1打印之前,thread2执行了三次打印计数。


程序:

#include <rtthread.h>

struct rt_thread thread1;
struct rt_thread thread2;

static rt_uint8_t thread1_stack[512];
static rt_uint8_t thread2_stack[512];

static void thread1_entry(void *parameter)
{
    static rt_uint32_t count = 0;

    for (; count<5; count++)
    {
        rt_thread_delay(RT_TICK_PER_SECOND * 3);
        
        rt_kprintf("count = %d\n", count);
    }
}

static void thread2_entry(void *parameter)
{
    rt_tick_t tick;
    rt_uint32_t i;

    for (i=0; i<15; i++)
    {
        tick = rt_tick_get();
        rt_thread_delay(RT_TICK_PER_SECOND);

        rt_kprintf("tick = %d\n", tick);
    }
}

int rt_application_init()
{
    rt_err_t result;

    result = rt_thread_init(&thread1,
        "t1",
        thread1_entry, RT_NULL, 
        &thread1_stack[0], sizeof(thread1_stack), 
        5, 5);

    if (result == RT_EOK)
        rt_thread_startup(&thread1);

    result = rt_thread_init(&thread2, 
        "t2",
        thread2_entry, RT_NULL, 
        &thread2_stack[0], sizeof(thread2_stack), 
        6, 5);

    if (result == RT_EOK)
        rt_thread_startup(&thread2);

    return 0;
}

/*@}*/  

                 

输出结果:

\ | /
- RT - Thread Operating System
/ | \ 1.1.0 build Aug 10 2012
2006 - 2012 Copyright by rt-thread team
tick = 1
tick = 1001
count = 0
tick = 2001
tick = 3002
tick = 4002
count = 1
tick = 5002
tick = 6002
tick = 7002
count = 2
tick = 8002
tick = 9002
tick = 10002
count = 3
tick = 11003
tick = 12004
tick = 13005
count = 4
tick = 14006


posted @ 2013-02-18 11:20  lyyyuna  阅读(264)  评论(0编辑  收藏  举报