[RTT例程练习] 1.3 线程让出

RTT 支持相同优先级，而ucosii 不支持。

如果一个线程不调用rt-thread_delay() 来让出调度器，那么它就会一直运行，其它线程永远处于就绪态。

而相同优先级的线程，在初始化或创建时还定义了其单次运行的最长的时间片，强迫其让出调度器。

这里，使用

rt_thread_yield();

也可让出调度器。

#include <rtthread.h>

rt_thread_t tid1 = RT_NULL;
rt_thread_t tid2 = RT_NULL;

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

while (1)
{
rt_kprintf(" thread1: count = %d\n", count++);

rt_thread_yield();
}
}

static void thread2_entry(void* parameter)
{
rt_uint32_t count = 0;

while (1)
{
rt_thread_yield();

rt_kprintf(" thread2: count = %d\n", count++);
}
}

/* 两者的优先级必须相同，否则一个线程让出调度器之后进入就绪队列，
其优先级仍比另一个高，会继续执行，使得另一个线程始终无法得到
运行。 */
int rt_application_init()
{
tid1 = rt_thread_create("t1",
thread1_entry,
RT_NULL,
512,
6, 10);
if (tid1 != RT_NULL)
rt_thread_startup(tid1);

tid2 = rt_thread_create("t2",
thread2_entry,
RT_NULL,
512,
6, 10);
if (tid2 != RT_NULL)
rt_thread_startup(tid2);

return 0;
}

/*@}*/

输出结果为：

\ | /
- RT - Thread Operating System
/ | \ 1.1.0 build Aug 10 2012
2006 - 2012 Copyright by rt-thread team
thread1: count = 0
thread1: count = 1
thread2: count = 0
thread1: count = 2
thread2: count = 1
thread1: count = 3
thread2: count = 2
thread1: count = 4
thread2: count = 3
thread2: count = 4