RTT软件仿真v0.1.0不能正常调度是何原因

#include <rtthread.h>
#define THREAD_PRIORITY 8
#define THREAD_TIMESLICE 5
/* 指 向 互 斥 量 的 指 针 */
static rt_mutex_t dynamic_mutex = RT_NULL;
static rt_uint8_t number1,number2 = 0;
ALIGN(RT_ALIGN_SIZE)
static char thread1_stack[1024];
static struct rt_thread thread1;
static void rt_thread_entry1(void parameter)
{
while(1)
{
/ 线 程 1 获 取 到 互 斥 量 后， 先 后 对 number1、number2 进 行 加 1 操 作， 然 后 释 放
互 斥 量 */
//rt_mutex_take(dynamic_mutex, RT_WAITING_FOREVER);
number1++;
rt_kprintf("num1加1操作\n");
rt_thread_mdelay(100);
rt_kprintf("num2加1操作\n");
number2++;
//rt_mutex_release(dynamic_mutex);
}
}
ALIGN(RT_ALIGN_SIZE)
static char thread2_stack[1024];
static struct rt_thread thread2;
static void rt_thread_entry2(void parameter)
{
while(1)
{
/ 线 程 2 获 取 到 互 斥 量 后， 检 查 number1、number2 的 值 是 否 相 同， 相 同 则 表 示
mutex 起 到 了 锁 的 作 用 /
rt_mutex_take(dynamic_mutex, RT_WAITING_FOREVER);
if(number1 != number2)
{
rt_kprintf("not protect.number1 = %d, mumber2 = %d \n",number1 ,number2)
;
}
else
{
rt_kprintf("mutex protect ,number1 = mumber2 is %d\n",number1);
}
number1++;
number2++;
rt_mutex_release(dynamic_mutex);
if(number1>=50)
return;
}
}
/ 互 斥 量 示 例 的 初 始 化 /
int mutex_sample(void)
{
/ 创 建 一 个 动 态 互 斥 量 /
dynamic_mutex = rt_mutex_create("dmutex", RT_IPC_FLAG_FIFO);
if (dynamic_mutex == RT_NULL)
{
rt_kprintf("create dynamic mutex failed.\n");
return -1;
}
rt_thread_init(&thread1,
"thread1",
rt_thread_entry1,
RT_NULL,
&thread1_stack[0],
sizeof(thread1_stack),
THREAD_PRIORITY-1, THREAD_TIMESLICE);
rt_thread_startup(&thread1);
rt_thread_init(&thread2,
"thread2",
rt_thread_entry2,
RT_NULL,
&thread2_stack[0],
sizeof(thread2_stack),
THREAD_PRIORITY, THREAD_TIMESLICE);
rt_thread_startup(&thread2);
return 0;
}
/ 导 出 到 MSH 命 令 列 表 中 */
MSH_CMD_EXPORT(mutex_sample, mutex_sample);