Priority-based scheduling in FreeRTOS - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of task execution order with different priorities

Consider three FreeRTOS tasks with priorities 2, 3, and 1 respectively. Each task prints its name once when it runs. What is the order of printed task names?

FreeRTOS

void Task1(void *pvParameters) {
    printf("Task1\n");
    vTaskDelete(NULL);
}
void Task2(void *pvParameters) {
    printf("Task2\n");
    vTaskDelete(NULL);
}
void Task3(void *pvParameters) {
    printf("Task3\n");
    vTaskDelete(NULL);
}

int main() {
    xTaskCreate(Task1, "Task1", 1000, NULL, 2, NULL);
    xTaskCreate(Task2, "Task2", 1000, NULL, 3, NULL);
    xTaskCreate(Task3, "Task3", 1000, NULL, 1, NULL);
    vTaskStartScheduler();
    return 0;
}

A Task1\nTask3\nTask2\n

B Task2\nTask1\nTask3\n

C Task3\nTask1\nTask2\n

D Task1\nTask2\nTask3\n

💡 Hint

Remember, FreeRTOS runs the highest priority task that is ready.

Task2 has the highest priority (3), so it runs first. Then Task1 (priority 2), then Task3 (priority 1).

❓ Predict Output

intermediate

Output when two tasks have the same priority

Two FreeRTOS tasks have the same priority and both are ready to run. Each prints its name once. What is the possible output order?

FreeRTOS

void TaskA(void *pvParameters) {
    printf("TaskA\n");
    vTaskDelete(NULL);
}
void TaskB(void *pvParameters) {
    printf("TaskB\n");
    vTaskDelete(NULL);
}

int main() {
    xTaskCreate(TaskA, "TaskA", 1000, NULL, 2, NULL);
    xTaskCreate(TaskB, "TaskB", 1000, NULL, 2, NULL);
    vTaskStartScheduler();
    return 0;
}

A No output, deadlock

B TaskA\nTaskB\n

C TaskB\nTaskA\n

D Either TaskA\nTaskB\n or TaskB\nTaskA\n

💡 Hint

Tasks with the same priority are scheduled round-robin.

When tasks have the same priority, FreeRTOS time-slices between them, so either task can run first.

🔧 Debug

advanced

Identify the cause of priority inversion

Given three tasks with priorities 3 (high), 2 (medium), and 1 (low). The low priority task holds a mutex needed by the high priority task. The medium priority task runs continuously. What problem occurs and why?

FreeRTOS

/*
TaskLow (priority 1) locks mutex and starts long processing
TaskHigh (priority 3) tries to lock mutex and blocks
TaskMedium (priority 2) runs continuously
*/

A No problem, TaskHigh preempts others immediately

B Deadlock because TaskHigh and TaskLow wait on each other

C Priority inversion because TaskMedium prevents TaskLow from running and releasing mutex

D Starvation of TaskMedium because TaskHigh blocks it

💡 Hint

Think about how medium priority tasks affect low priority tasks holding resources needed by high priority tasks.

Priority inversion happens because the medium priority task keeps running, preventing the low priority task from releasing the mutex needed by the high priority task.

📝 Syntax

advanced

Identify the syntax error in task creation

Which option contains the correct syntax to create a FreeRTOS task with priority 4?

A xTaskCreate(TaskFunc, "Task", 1000, NULL, 4, NULL);

B xTaskCreate(TaskFunc, Task, 1000, NULL, 4, NULL);

C xTaskCreate(TaskFunc, "Task", 1000, NULL, "4", NULL);

D xTaskCreate(TaskFunc, "Task", 1000, NULL, 4);

💡 Hint

Check the parameter types and count for xTaskCreate.

Option A uses correct string for task name and integer for priority with all required parameters.

🚀 Application

expert

Determine the number of context switches in priority scheduling

In a FreeRTOS system, three tasks with priorities 3, 2, and 1 are ready. The highest priority task runs for 5 ticks, then blocks. The next highest runs for 3 ticks, then blocks. The lowest priority task runs for 4 ticks. How many context switches occur in total?

A 4

B 3

C 5

D 6

💡 Hint

Count switches when tasks block and scheduler switches to next ready task.

Initially, highest priority runs (no switch). When it blocks, switch to medium priority (1). When medium blocks, switch to low priority (2). When low finishes, no more tasks, so total 2 switches. But starting scheduler counts as first switch (0 to highest), total 3. Also, when tasks delete themselves, scheduler switches back to idle task (3). Total 4 context switches.