Task notification vs queue performance in FreeRTOS - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to send a task notification to unblock the task.

FreeRTOS

xTaskNotifyGive([1]);

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AxSemaphore

BxTaskNotify

CxQueueHandle

DxTaskHandle

Attempts:

3 left

💡 Hint

Common Mistakes

Using a queue handle instead of a task handle.

Confusing the notification function with queue send functions.

The function xTaskNotifyGive() requires the handle of the task to notify, which is xTaskHandle.

2fill in blank

medium

Complete the code to receive a notification and unblock the task.

FreeRTOS

ulTaskNotifyTake([1], portMAX_DELAY);

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ApdTRUE

BpdFALSE

CpdFAIL

DpdPASS

Attempts:

3 left

💡 Hint

Common Mistakes

Passing pdFALSE which does not clear the notification value.

Using unrelated constants like pdPASS or pdFAIL.

The first parameter of ulTaskNotifyTake() is pdTRUE to clear the notification value on exit.

3fill in blank

hard

Fix the error in the queue send function to avoid blocking the task indefinitely.

FreeRTOS

xQueueSend([1], &item, 0);

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AxQueue

BxTaskHandle

CxSemaphore

DxNotification

Attempts:

3 left

💡 Hint

Common Mistakes

Using a task handle instead of a queue handle.

Confusing notification handles with queue handles.

The first parameter of xQueueSend() must be a queue handle, here xQueue.

4fill in blank

hard

Fill both blanks to create a dictionary mapping task names to their notification counts, filtering tasks with counts greater than 5.

FreeRTOS

task_notifications = {task_name: [1] for task_name, count in tasks.items() if count [2] 5}

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Acount

B>

C<

Dtask_name

Attempts:

3 left

💡 Hint

Common Mistakes

Using task names as values instead of counts.

Using the wrong comparison operator.

The dictionary comprehension maps task_name to count for counts greater than 5.

5fill in blank

hard

Fill all three blanks to create a dictionary of tasks with notification counts above zero, using uppercase task names as keys.

FreeRTOS

active_tasks = {{ [1]: [2] for [3], count in tasks.items() if count > 0 }}

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Atask_name.upper()

Bcount

Ctask_name

Dcount > 0

Attempts:

3 left

💡 Hint

Common Mistakes

Using the condition inside the dictionary instead of the comprehension filter.

Mixing up keys and values.

The dictionary keys are uppercase task names, values are counts, iterating over task_name and count.