Richard,

I'm making videos of FreeRTOS tutorial (http://www.socialledge.com/sjsu/index.php?title=FreeRTOS_Tutorial) and I've noticed the priority inversion anomaly in FreeRTOS. In my demo program below, it makes no difference if the semaphore is a mutex or a binary semaphore. In either case, I see the same output, but I think it lower priority tasks should get equal access.

Output (regardless if semaphore is a mutex or semaphore):
task4 got sem. Will use it ...
task3 got sem. Will use it ...
task4 got sem. Will use it ...
task3 got sem. Will use it ...
task4 got sem. Will use it ...

xSemaphoreHandle sd_card_lock = 0;
void my_task(void *p)
{
while(1) {
if(xSemaphoreTake(sd_card_lock, 999999)) {
printf("%s got sem. Will use it ... \n", (char*)p);
xSemaphoreGive(sd_card_lock);
}
vTaskDelay(1);
}
}
void main( void )
{
sd_card_lock = xSemaphoreCreateMutex();
//vSemaphoreCreateBinary(sd_card_lock);

xTaskCreate(my_task, (signed char*) "t1", 1024, (void*)"task1", 2, NULL );
xTaskCreate(my_task, (signed char*) "t2", 1024, (void*)"task2", 2, NULL );
xTaskCreate(my_task, (signed char*) "t3", 1024, (void*)"task3", 3, NULL );
xTaskCreate(my_task, (signed char*) "t4", 1024, (void*)"task4", 3, NULL );
vTaskStartScheduler();
}

If all tasks have the same priority, then we see every task taking turn, again regardless if the semaphore is a mutex or binary. But with a mutex, it should be the same behavior but it's not :( Can you please educate me?

Preet

Why would you expect the priority 1 or 2 task to every get the semaphore?

When Task 4 gets the semaphore, everyone else lines up on it and waits for it.
When Task 4 gives the semaphore back, Task3, since it is the highest priority task will get it.
While Task 3 is outputing its message, my guess is that Task 4 finishes its 1 tick wait, and then puts itself in line for the semaphore.
When Task 3 finishes, task 4 get the semaphore because it is the highest priority task waiting on the semaphore.

Making the semaphore into a mutex just says that when Task 4 starts its wait on the semaphore, Task 3 gets a temporary boost in priority to that of Task 4 (the priority inversion logic) .

If all tasks are the same priority, then the priority doesn't select which task gets chosen, and among equal priority tasks, the one waiting the longest it chosen first.