RTOS mechanics when there is a main task and a second task

Assuming that you have a main task and a second task thus.


main task 0
start task 1
do something
time delay 20 ticks

Task 1
x=getkey or some loop waiting for input
time delay 5 ticks

Now the question is, in task1 as it is in effect in a loop waiting for input, it may never get to the delay tick routine. Does this mean that it is switched in every 5 ticks, or in fact only when it gets an input. Or indeed is it switched at the rtos default setting, and the 5 ticks has no effect until reached. Not really sure about the mechanics of operation. anyone clued up on this?

Re: RTOS question

Whenever any task call delay ticks, it will be blocked and relinguish CPU to other task.

main task 0
start task 1
do something
time delay 20 ticks // relinquish CPU, task1 will get executed

Task 1
x=getkey or some input // loop waiting is not good pratice in RTOS
time delay 5 ticks // task 1 blocked for 5 ticks, task 0 will get executed

Re: RTOS question

When Task 1 begins executions and i made to wait for any resources like input from user and delay , Scheduler pre empts or pends that task which i wating for any resources other then CPU so the executions to task ) or what ever task and resumes when task 1 is ready

RTOS question

I agree with Epegic

main task 0
start task 1
do something
time delay 20 ticks // relinquish CPU, task1 will get executed

Task 1
x=getkey or some input // loop waiting is not good pratice in RTOS
time delay 5 ticks // task 1 blocked for 5 ticks, task 0 will get executed

Re: RTOS question

Yes, you are correct when you state that Task 1 will not reach the delay tick routine until it gets an input. Therefore, the "delay 5 ticks" statement has no effect until it is reached.

If the "getkey" routine suspends waiting for a key, Task 1 will of course be switched out when it calls getkey. When it receives a key, it will be switched in.

If Task 1 is in a loop waiting for a key, it will not be switched out....unless Main Task 0 is higher priority. Then, Main Task 0 will pre-empt Task 1 every 20 ticks.

Re: RTOS question

GrandAlf said:

Assuming that you have a main task and a second task thus.


main task 0
start task 1
do something
time delay 20 ticks

Task 1
x=getkey or some loop waiting for input
time delay 5 ticks

Now the question is, in task1 as it is in effect in a loop waiting for input, it may never get to the delay tick routine. Does this mean that it is switched in every 5 ticks, or in fact only when it gets an input. Or indeed is it switched at the rtos default setting, and the 5 ticks has no effect until reached. Not really sure about the mechanics of operation. anyone clued up on this?

Click to expand...

To me it looks like algo of main_task_0. So main_task_0 spawns task_1. When it does this, the RTOS scheduler will create a new task called task_1 and it will become ready for execution. Now we have three cases:

Case 1:
if the priority of main_task_0 is greater than task_1, then it will continue to execute. If "do something" does not cause it to block or pend, then it shall reach time_delay statement where it shall be switched out and task_1 shall start executing. In task_1, input is detecte in busy-waiting loop. So it shall continue to execute till either it gets the input or main_task_0 wakes up. Supposing task_1 gets input first, then it shall again pend due to time_delay statement. At this point, system idle task shall be running and system shall be doing no useful work.

Case 2:
Assuming that priority of the task_1 is greater than main_task_0, then in this case the main_task_0 won't be able to complete the "do-something" part of its work and task_1 shall take over. The system shall continue to wait for input in a busy loop and once input is received, the task_1 executes time_delay to hand over CPU back to main_task_0 which continues with "do something". If this "do something" part is completed before 5 ticks, then the main_task_0 surrenders CPU via time_delay and system becomes idle.

Case 3:
Assuming both tasks have same priority, then it depends. Most likely, RTOS shall allow the current executing task to continue. If Round-Robin scheduling for equal priority task is enabled then probably both tasks shall execute for roughly equal amount of the time - task_1 executing the busy loop while main_task_0 executing the "do something" portion of its task.