FreeRTOS Mutex Unexpected behaviour - freertos

Question

#include "main.h"
#include "FreeRTOS.h"
#include "semphr.h"
#include "task.h"
/* Private function prototypes -----------------------------------------------*/
void SystemClock_Config(void);
static void MX_GPIO_Init(void);
static void MX_USART2_UART_Init(void);
void Task1(void *argument);
void Task2(void *argument);
void PrintMsg(char *data);
/* USER CODE BEGIN PFP */
SemaphoreHandle_t hMutex = NULL;
TaskHandle_t hTask1 = NULL;
TaskHandle_t hTask2 = NULL;
UART_HandleTypeDef huart2;
int main(void) {
/* MCU Configuration--------------------------------------------------------*/
/* Reset of all peripherals, Initializes the Flash interface and the Systick. */
HAL_Init();
/* Configure the system clock */
SystemClock_Config();
/* Initialize all configured peripherals */
MX_GPIO_Init();
MX_USART2_UART_Init();
/* USER CODE BEGIN 2 */
xTaskCreate(Task1, "Task 1", configMINIMAL_STACK_SIZE, NULL, 2, &hTask1);
xTaskCreate(Task2, "Task 2", configMINIMAL_STACK_SIZE, NULL, 2, &hTask2);
/* USER CODE END 2 */
hMutex = xSemaphoreCreateMutex();
if (hMutex == NULL) {
PrintMsg("Mutex not created\r\n");
} else {
PrintMsg("Mutex created\r\n");
}
vTaskStartScheduler();
/* USER CODE BEGIN WHILE */
while (1) {
/* USER CODE END WHILE */
/* USER CODE BEGIN 3 */
}
/* USER CODE END 3 */
}
void Task1(void *argument) {
/* USER CODE BEGIN 5 */
/* Infinite loop */
for (;;) {
if (xSemaphoreTake(hMutex,2000) == pdTRUE) {
PrintMsg("Shared Resource Start and Executing Task 1\r\n");
xSemaphoreGive(hMutex);
PrintMsg("Shared Resource End and Executing Task 1\r\n");
vTaskDelay(100);
} else {
PrintMsg("Task 1 Didn't get access to shared resource\r\n");
}
}
/* USER CODE END 5 */
}
void Task2(void *argument) {
/* USER CODE BEGIN 5 */
/* Infinite loop */
for (;;) {
if (xSemaphoreTake(hMutex,2000) == pdTRUE) {
PrintMsg("Shared Resource Start and Executing Task 2\r\n");
//xSemaphoreGive(hMutex);
PrintMsg("Shared Resource End and Executing Task 2\r\n");
vTaskDelay(100);
} else {
PrintMsg("Task 2 Didn't get access to shared resource\r\n");
}
}
/* USER CODE END 5 */
}
void PrintMsg(char *data) {
int i = 0;
while (*(data + i) != '\0') {
HAL_UART_Transmit(&huart2, (uint8_t*) &data[i], 1, 0xFF);
i++;
}
}
HW USED:STM32F446RE
When i run this code,I get output as follows
//Output Start
Mutex created
Shared Resource Start and Executing Task 1
Shared Resource End and Executing Task 1
Task 2 Didn't get access to shared resource
Task 1 DiTask 2 Didn't get access to shared resource
Task 1 Didn't get access to shared resource
Task 2 Didn't get access to shared resource
.
.
//Output End
Question1)
Considering Task1 was scheduled first as the priority of Task1 and 2 are same.
Task1 is executed correctly.
After this,Task2 is scheduled but was not able to take the mutex because of which i got output as "Task 2 Didn't get access to shared resource".Why is this so?
Question2 )
"Task 1 DiTask 2 Didn't get access to shared resource" in this line,it seems that Task1 was exceuting but it got preempted by Task2 which should have not happened as both the Tasks have same priority??

Answer 1

Q1: The likely explanation here is that you're being fooled by competing tasks overwriting each others' buffered UART output. UART is slow, task switching is fast :)
After Task1 prints "Shared Resource Start and Executing Task 1" and releases the mutex for the first time, scheduler immediately switches to Task2 (maybe because Task1 has used up its time slot?). Task2 grabs the mutex, quickly spits both of its messages into UART buffer and enters sleep for 100 ticks. The scheduler immediately switches back to Task1, but unfortunately the buffered messages from Task2 haven't hit the UART output yet. UART buffer gets overwritten by the message "Shared Resource End and Executing Task 1" before a single byte of Task2-s messages is printed.
After that the entire process stalls because you haven't released the mutex in Task2.
Q2: tasks with equal priorities get preempted by each other when the scheduler decides so (usually when they enter sleep or have exhausted their time slot)
I would recommend this tutorial/book, chapter 3 for a great explanation of how tasks work.

Answer 2

USART is a resource, so it can't be shared by two tasks without a sync mechanism, like a mutex. Your usage of PrintMsg() violates this rule.
The scheduler is run at configTICK_RATE_HZ frequency. If configUSE_TIME_SLICING is 1 (or undefined), the scheduler switches between equal priority tasks. If you don't want this default behavior, set configUSE_TIME_SLICING to 0.
Keep it in mind that your configTICK_RATE_HZ setting of 1000 gives at most ~1 ms runtime to each task, unless there are no other tasks ready to run. Higher priority tasks can also preempt it before that 1 ms passes. With 115200 baud rate, you can send ~10 bytes or so in this ~1 ms time period.