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Scheduler behavior in non-preemptive mode

Posted by Tzu-Chien Chiu on August 29, 2006
I have a non-preemptive port of FreeRTOS and there there are totally three tasks:

Task 1, tskIDLE_PRIORITY + 1
Task 2, tskIDLE_PRIORITY + 2

Task 1 and 2 is created using the same function:

portTASK_FUNCTION(task_proc, params)
unsigned id = *(unsigned*) params;
while (1) {
printf("*%d* ", id);

But Task 1 *never* gets a chance to run. After reading vTaskSwitchContext(), I found that Task 1 is starved because Task 2 is *always* in the ready list, and uxTopReadyPriority is always 2.

Is this behavior expected?

Tzu-Chien Chiu - SMedia Technology Corp.
URL: http://www.csie.nctu.edu.tw/~jwchiu/

RE: Scheduler behavior in non-preemptive mode

Posted by Nobody/Anonymous on August 29, 2006
Yes, this would be expected because task 2 never blocks. It is always the highest priority task that is able to run.

If you replaced taskYIELD() with vTaskDelay( 2 ), then task 2 would block for 2 ticks, allowing task 1 to execute.

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