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