您的位置:首页 > 其它

[RTT例程练习] 1.5 优先级相同线程轮转调度

2013-02-18 11:27 447 查看
之前说过,相同优先级的线程,在自己的时间片用光之后,会被剥夺调度器,让给同优先级的其他线程。

程序:

#include <rtthread.h>

static struct rt_thread thread1;
static struct rt_thread thread2;
static char thread1_stack[512];
static char thread2_stack[512];

//static rt_uint32_t t1_count = 0;
//static rt_uint32_t t2_count = 0;
static void thread1_entry(void* parameter)
{
rt_uint8_t i;
for(i = 0; i < 6; i ++)
{
rt_kprintf("Thread11111111111111111111111111:\n\r");
rt_kprintf("This is \n");
rt_kprintf("a\n");
rt_kprintf("demo\n");
rt_thread_delay(10);
}
}

static void thread2_entry(void* parameter)
{
rt_uint8_t j;
for(j = 0; j <60; j ++)
{
rt_kprintf("Thread2:\n\r");
rt_kprintf("This is \n");
rt_kprintf("a\n");
rt_kprintf("demo\n");
}
}

int rt_application_init()
{
rt_err_t result;

result = rt_thread_init(&thread1,
"t1",
thread1_entry, RT_NULL,
&thread1_stack[0], sizeof(thread1_stack),
6, 10);
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 result;
}

/*@}*/


输出结果:
Thread1:
This is
a
demo
Thread2:
This is
a
demo
Thread2:
This is
a
demo
Thread2:
This is
a
demo
Thread2:
This is
a
demo
Thread2:
This is
a
demo
Thread2:
This isThread11111111111111111111111111:
This is
a
demo

a
demo
Thread2:
This is
a
demo
Thread2:
This is
a
demo
Thread2:
This is
a
demo
Thread2:
This is
a
Demo


可以看到thread2 的时间片用光,插入了thread1 的打印内容。
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: