Why mutexes protect shared resources in FreeRTOS - See It in Action

Why mutexes protect shared resources

📖 Scenario: Imagine two tasks in a FreeRTOS system that both want to update the same counter variable. Without control, they might change it at the same time, causing errors.

🎯 Goal: Build a simple FreeRTOS program where two tasks safely increment a shared counter using a mutex to avoid conflicts.

📋 What You'll Learn

Create a shared variable called counter initialized to 0

Create a mutex handle called xMutex

Write two tasks named vTask1 and vTask2 that each increment counter 5 times

Use xSemaphoreTake and xSemaphoreGive with xMutex to protect the increments

Print the value of counter after each increment

💡 Why This Matters

🌍 Real World

In embedded systems, multiple tasks often share hardware or data. Mutexes keep these accesses safe and predictable.

💼 Career

Understanding mutexes is essential for embedded software engineers working with real-time operating systems like FreeRTOS.

1

Create the shared counter variable

Create a global integer variable called counter and set it to 0.

FreeRTOS

# Create a global integer variable called counter and set it to 0
# Your code here

Hint

Use int counter = 0; outside any function to create the shared variable.

2

Create the mutex handle

Create a global variable called xMutex of type SemaphoreHandle_t and initialize it by calling xSemaphoreCreateMutex() inside main().

FreeRTOS

int counter = 0;

// Create a mutex handle called xMutex and initialize it in main
// Your code here

Hint

Declare SemaphoreHandle_t xMutex; globally and call xSemaphoreCreateMutex() in main().

3

Write two tasks that increment the counter safely

Write two FreeRTOS tasks named vTask1 and vTask2. Each task should loop 5 times, and inside each loop, use xSemaphoreTake(xMutex, portMAX_DELAY) before incrementing counter by 1, then call xSemaphoreGive(xMutex). Use vTaskDelay(10) after releasing the mutex.

FreeRTOS

int counter = 0;

SemaphoreHandle_t xMutex;

int main(void) {
    xMutex = xSemaphoreCreateMutex();
    // Create tasks vTask1 and vTask2 here
    // Your code here
    return 0;
}

// Define vTask1 and vTask2 functions below
// Your code here

Hint

Use xSemaphoreTake and xSemaphoreGive around the increment to protect counter.

4

Print the counter value after each increment

Inside both vTask1 and vTask2, after incrementing counter and before releasing the mutex, add a printf statement to print the current value of counter with the format: "Counter value: %d\n".

FreeRTOS

int counter = 0;

SemaphoreHandle_t xMutex;

void vTask1(void *pvParameters) {
    for (int i = 0; i < 5; i++) {
        if (xSemaphoreTake(xMutex, portMAX_DELAY) == pdTRUE) {
            counter++;
            // Print counter value here
            // Your code here
            xSemaphoreGive(xMutex);
            vTaskDelay(10);
        }
    }
    vTaskDelete(NULL);
}

void vTask2(void *pvParameters) {
    for (int i = 0; i < 5; i++) {
        if (xSemaphoreTake(xMutex, portMAX_DELAY) == pdTRUE) {
            counter++;
            // Print counter value here
            // Your code here
            xSemaphoreGive(xMutex);
            vTaskDelay(10);
        }
    }
    vTaskDelete(NULL);
}

int main(void) {
    xMutex = xSemaphoreCreateMutex();
    xTaskCreate(vTask1, "Task 1", 1000, NULL, 1, NULL);
    xTaskCreate(vTask2, "Task 2", 1000, NULL, 1, NULL);
    vTaskStartScheduler();
    for(;;);
}

Hint

Use printf("Counter value: %d\n", counter); inside the mutex protection block.

Practice

1. Why do we use mutexes in FreeRTOS when accessing shared resources?

easy

A. To allow only one task to access the resource at a time

B. To speed up the execution of all tasks simultaneously

C. To permanently lock the resource so no task can use it

D. To create multiple copies of the resource for each task

Solution

1. Step 1: Understand what a shared resource is

   A shared resource is something multiple tasks want to use, like a variable or hardware.

2. Step 2: Understand mutex purpose in FreeRTOS

   Mutexes ensure only one task uses the resource at a time to avoid conflicts or errors.

3. Final Answer:

   To allow only one task to access the resource at a time -> Option A

4. Quick Check:

   Mutex = single task access [OK]

Hint: Mutex means one task at a time [OK]

Common Mistakes:

• Thinking mutex speeds up all tasks
• Believing mutex locks resource forever
• Assuming mutex duplicates resources

2. Which FreeRTOS function is used to lock a mutex before accessing a shared resource?

easy

A. xSemaphoreGive()

B. vTaskDelay()

C. xSemaphoreTake()

D. xQueueSend()

Solution

1. Step 1: Identify the function to lock a mutex

   In FreeRTOS, xSemaphoreTake() is used to take or lock a mutex.

2. Step 2: Confirm other functions' purposes

   xSemaphoreGive() releases the mutex, vTaskDelay() delays a task, and xQueueSend() sends data to a queue.

3. Final Answer:

   xSemaphoreTake() -> Option C

4. Quick Check:

   Lock mutex = xSemaphoreTake() [OK]

Hint: Take locks, Give unlocks mutex [OK]

Common Mistakes:

• Confusing xSemaphoreGive() as lock
• Using vTaskDelay() to lock mutex
• Mixing queue functions with mutex

3. What will happen if two tasks try to access a shared variable without using a mutex in FreeRTOS?

medium

A. Data corruption or unexpected behavior may occur

B. Both tasks access safely without issues

C. The system automatically locks the variable

D. The second task waits automatically until the first finishes

Solution

1. Step 1: Understand shared resource risks without mutex

   Without mutex, tasks can change the variable at the same time, causing data corruption.

2. Step 2: Check FreeRTOS behavior without explicit locking

   FreeRTOS does not automatically lock variables; tasks run concurrently unless synchronized.

3. Final Answer:

   Data corruption or unexpected behavior may occur -> Option A

4. Quick Check:

   No mutex = risk of data corruption [OK]

Hint: No mutex means unsafe shared access [OK]

Common Mistakes:

• Assuming FreeRTOS auto-locks variables
• Thinking tasks wait automatically
• Believing no issues happen without mutex

4. Given this code snippet, what is the error related to mutex usage?

if(xSemaphoreTake(mutex, 0) == pdTRUE) {
  sharedVar++;
}
// Missing xSemaphoreGive(mutex) here

medium

A. The mutex is taken twice without giving

B. The mutex is given before taking

C. The shared variable is not incremented

D. The mutex is never released, causing a deadlock

Solution

1. Step 1: Check mutex take and give usage

   The code takes the mutex but never calls xSemaphoreGive() to release it.

2. Step 2: Understand consequences of not releasing mutex

   Not releasing the mutex causes other tasks to wait forever, leading to deadlock.

3. Final Answer:

   The mutex is never released, causing a deadlock -> Option D

4. Quick Check:

   Missing give = deadlock risk [OK]

Hint: Always give mutex after take [OK]

Common Mistakes:

• Forgetting to release mutex
• Assuming increment fails without mutex
• Confusing order of take/give

5. You have two tasks that both update a shared sensor reading variable. How should you use a mutex to ensure safe updates and avoid blocking tasks too long?

hard

A. Take the mutex once at system start and never release it

B. Take the mutex, update the variable quickly, then give the mutex immediately

C. Update the variable without mutex and rely on task priorities

D. Use two separate mutexes for the same variable

Solution

1. Step 1: Understand mutex best practice for shared data

   Mutex should be held only briefly to protect the critical section (variable update).

2. Step 2: Evaluate options for safe and efficient updates

   Taking and giving mutex immediately around update prevents blocking other tasks too long.

3. Final Answer:

   Take the mutex, update the variable quickly, then give the mutex immediately -> Option B

4. Quick Check:

   Short mutex hold = safe and efficient [OK]

Hint: Lock only during update, then unlock fast [OK]

Common Mistakes:

• Holding mutex too long blocking tasks
• Never releasing mutex causing deadlock
• Ignoring mutex and risking data errors