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分享关于vxworks轮转调度
下面是《vxworks与tornado》的例子：
#define PRIORITY 50
#define TIMESFLICE (float)sysClkRateGet()/100
void sched(void)
{
int taskIdOne, taskIdTwo, taskIdThree;
int i, j;
if(kernelTimeSlice(TIMESLICE) == OK)
printf("\n\n\n\n\t\t\tTIMESLICE = %f secondes\n\n\n", TIMESLICE/60);
else
printf("\n\n\n\n\t\t\tTIMESLICE SET ERROR!\n\n\n");
if((taskIdOne =
taskSpawn("task1", PRIORITY, 0x100, 2000, taskOne,
0,0,0,0,0,0,0,0,0,0)) == ERROR);
printf("taskSpawn taskOne failed\n");
if((taskIdTwo =
taskSpawn("task2", PRIORITY, 0x100, 2000, taskTwo,
0,0,0,0,0,0,0,0,0,0)) == ERROR);
printf("taskSpawn taskTwo failed\n");
if((taskIdThree =
taskSpawn("task3", PRIORITY, 0x100, 2000, taskThree,
0,0,0,0,0,0,0,0,0,0)) == ERROR);
printf("taskSpawn taskThree failed\n");
for(i = 0;i < ITER1;i++)
{
for(j = 0; j < ITER2; j++)
logMsg("\n", 0, 0, 0, 0, 0, 0);
for(j = 0; j < LONG_TIME; j++);
}
}
sched的优先级为100,taskSpawn出来的任务的优先级都是50，实现轮转调度，发现放在sched代码后的循环语句还是在新建的三个任务前运行。
轮转调度不是先运行高优先级的任务吗？以为sched中的循环语句应该在所有新建任务完成后在执行的，请高手指教。
#define PRIORITY 50
#define TIMESFLICE (float)sysClkRateGet()/100
void sched(void)
{
int taskIdOne, taskIdTwo, taskIdThree;
int i, j;
if(kernelTimeSlice(TIMESLICE) == OK)
printf("\n\n\n\n\t\t\tTIMESLICE = %f secondes\n\n\n", TIMESLICE/60);
else
printf("\n\n\n\n\t\t\tTIMESLICE SET ERROR!\n\n\n");
if((taskIdOne =
taskSpawn("task1", PRIORITY, 0x100, 2000, taskOne,
0,0,0,0,0,0,0,0,0,0)) == ERROR);
printf("taskSpawn taskOne failed\n");
if((taskIdTwo =
taskSpawn("task2", PRIORITY, 0x100, 2000, taskTwo,
0,0,0,0,0,0,0,0,0,0)) == ERROR);
printf("taskSpawn taskTwo failed\n");
if((taskIdThree =
taskSpawn("task3", PRIORITY, 0x100, 2000, taskThree,
0,0,0,0,0,0,0,0,0,0)) == ERROR);
printf("taskSpawn taskThree failed\n");
for(i = 0;i < ITER1;i++)
{
for(j = 0; j < ITER2; j++)
logMsg("\n", 0, 0, 0, 0, 0, 0);
for(j = 0; j < LONG_TIME; j++);
}
}
sched的优先级为100,taskSpawn出来的任务的优先级都是50，实现轮转调度，发现放在sched代码后的循环语句还是在新建的三个任务前运行。
轮转调度不是先运行高优先级的任务吗？以为sched中的循环语句应该在所有新建任务完成后在执行的，请高手指教。
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ies_sweet 2009-01-20
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顶贴，学习中，愿高手指点迷津。
freeman11me 2009-01-10
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这个代码运行起来一直没有看到理论中的那种task1,task2,task3轮流运行一段时间的效果：
运行结果如下：
而且sched在执行taskspawn的时候，还出现了taskspawn failed错误，既然taskspawn错误了，为何最终3个还是生成了、执行了？
#include "vxWorks.h"
#include "taskLib.h"
#include "kernelLib.h"
#include "sysLib.h"
#include "logLib.h"
void taskOne(void);
void taskTwo(void);
void taskThree(void);
#define ITER1 5
#define ITER2 10
#define PRIORITY 50
#define TIMESLICE (float)sysClkRateGet()/100
#define LONG_TIME 1000000
void sched(void)
{
int taskIdOne, taskIdTwo, taskIdThree;
int i, j;
if(kernelTimeSlice(TIMESLICE) == OK)
printf("\n\n\n\n\t\t\tTIMESLICE = %f secondes\n\n\n", TIMESLICE/60);
else
printf("\n\n\n\n\t\t\tTIMESLICE SET ERROR!\n\n\n");
if((taskIdOne =
taskSpawn("task1", PRIORITY, 0x100, 2000, taskOne,
0,0,0,0,0,0,0,0,0,0)) == ERROR);
printf("taskSpawn taskOne failed\n");
if((taskIdTwo =
taskSpawn("task2", PRIORITY, 0x100, 2000, taskTwo,
0,0,0,0,0,0,0,0,0,0)) == ERROR);
printf("taskSpawn taskTwo failed\n");
if((taskIdThree =
taskSpawn("task3", PRIORITY, 0x100, 2000, taskThree,
0,0,0,0,0,0,0,0,0,0)) == ERROR);
printf("taskSpawn taskThree failed\n");
}
void taskOne(void)
{
int i, j;
for(i = 0;i < ITER1;i++)
{
for(j = 0; j < ITER2; j++)
logMsg("\n", 0, 0, 0, 0, 0, 0);
for(j = 0; j < LONG_TIME; j++);
}
}
void taskTwo(void)
{
int i, j;
for(i = 0;i < ITER1;i++)
{
for(j = 0; j < ITER2; j++)
logMsg("\n", 0, 0, 0, 0, 0, 0);
for(j = 0; j < LONG_TIME; j++);
}
}
void taskThree(void)
{
int i, j;
for(i = 0;i < ITER1;i++)
{
for(j = 0; j < ITER2; j++)
logMsg("\n", 0, 0, 0, 0, 0, 0);
for(j = 0; j < LONG_TIME; j++);
}
}
运行结果如下：
TIMESLICE = 0.010000 secondes
taskSpawn taskOne failed
taskSpawn taskTwo failed
taskSpawn taskThree failed
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3868de8 (task1):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x3865268 (task2):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
0x38616e8 (task3):
而且sched在执行taskspawn的时候，还出现了taskspawn failed错误，既然taskspawn错误了，为何最终3个还是生成了、执行了？
erigido 2009-01-04
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taskSpawn就是一个任务切换点，肯定是进行了切换。我也有同样的问题，task里面里面做什么了？
上面高手说的没错，logMsg为了保证在中断中可以调用，都是同一发给tLogTask任务来进行输出的。
最简单的方式，可以在内存中记录当前的cputick值看看。
上面高手说的没错，logMsg为了保证在中断中可以调用，都是同一发给tLogTask任务来进行输出的。
最简单的方式，可以在内存中记录当前的cputick值看看。
土豆 2008-12-31
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有两个问题：
1.你怎么确定“放在sched代码后的循环语句还是在新建的三个任务前运行”？是在你创建的task里面加打印?
2.logMsg本身通过消息队列实现，也就是说对logMsg的调用本身就是通过向另外一个任务tLogTask发送消息，通过tLogTask输出到屏幕的。
有时侯看到的并不是真实的。
1.你怎么确定“放在sched代码后的循环语句还是在新建的三个任务前运行”？是在你创建的task里面加打印?
2.logMsg本身通过消息队列实现，也就是说对logMsg的调用本身就是通过向另外一个任务tLogTask发送消息，通过tLogTask输出到屏幕的。
有时侯看到的并不是真实的。
