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A lower priority task starting first ?

Posted by Diptopal Basu on August 22, 2013
Here is a relevant part of the code

xSemaphoreHandle employee_signal = 0;

void employee_task() {

;

}

void boss(void *p){
for(;;){
LPC_printf("1\n\r");
xSemaphoreGive(employee_signal);
LPC_printf("3\r\n");
vTaskDelay(2000);

}
}


void employee(void *p){
for(;;){
if(xSemaphoreTake(employee_signal, portMAX_DELAY) == pdTRUE ){
employee_task();
LPC_printf("2\n\r");
}
}

}



int main (void) {

vSemaphoreCreateBinary(employee_signal);
LPC_UART0->IER = IER_THRE | IER_RLS;
SystemInit(); /* initialize clocks */
UARTInit(0, 38400);/* baud rate setting */
UARTInit(1, 38400);/* baud rate setting */
xTaskCreate( boss, ( signed char* ) "t1", 1024, NULL, 1, NULL );
xTaskCreate( employee, ( signed char* ) "t2", 1024, NULL, 1, NULL );
vTaskStartScheduler();
return 0;
}



I am expecting the boss task to start printing first, but as I can see, once immediately after the program starts the employee task is printing once to the console, then it falls into the right sequence. I am expecting the boss task to start instead, give a semaphore to the employee task and then the employee task to wake up. Pasting a part of the output here. This is on LPC1768 hardware, I hope I have configured the RTOS properly.


2 <- This should not be here
1 <- The printing should start from here
2
3
1
2
3
1
2
3


I am only able to avoid this situation id I am making the priority od the boss task higher. Then I am getting the correct output. How is the employee task printing first when the boss has not even started and given it a semaphore?

RE: A lower priority task starting first ?

Posted by Diptopal Basu on August 22, 2013
I am sorry, the topic should have been something else, like how can my task print even without getting a Semaphore?

RE: A lower priority task starting first ?

Posted by MEdwards on August 22, 2013
The semaphore is created so the first take will pass. If you take the semaphore immediately after you create it (before your tasks start using it) then I think it will work the way your code expected it to.

RE: A lower priority task starting first ?

Posted by Matthew Kendall on August 22, 2013
Mutexes and binary semaphores are created in the already-given state, so the fact that your first take operation succeeds is expected behaviour.

For it to work as you want, you must take the semaphore immediately after creating it.


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