FreeRTOS队列

Question

我现在还不熟悉FreeRTOS和学习队列。我有一个将5个整数发送到队列的Task_3和Task4，它读取队列并打印读取的值。问题是我将0..4写入队列，但Task 4读取队列的所有5个位置的值5。为什么不是0..4？

发送任务3

    void task_3(void *args)
    {
        uint32_t count = 0;
        uint32_t *Value;
    
        Value = &count;
        ESP_EARLY_LOGI(TAG, "Task 3 entered..");
    
        
        if (myQueue_2 == NULL)
        {
            printf("myQueue_2 could not be created !\n");
        }
        
        ESP_EARLY_LOGI("TASK 3", "Data waiting in myQueue_2_1 : %d, Space available : %d", uxQueueMessagesWaiting(myQueue_2), uxQueueSpacesAvailable(myQueue_2));
       
        for (count = 0; count < 5;count++)
        {
            ESP_EARLY_LOGI("TASK 3", "Value = %d\n", *Value);
            xQueueSend(myQueue_2, (void*)&Value, (TickType_t)0);
        }
    
        ESP_EARLY_LOGI("TASK 3", "Data waiting in myQueue_2_1 : %d, Space available : %d", 
        uxQueueMessagesWaiting(myQueue_2), uxQueueSpacesAvailable(myQueue_2));
        while (1)
        {
            vTaskDelay(pdMS_TO_TICKS(500));
        }
    }

任务3的产出

I (0) TASK 3: Data waiting in myQueue_2_1 : 0, Space available : 5
I (10) TASK 3: Value = 0
I (10) TASK 3: Value = 1
I (20) TASK 3: Value = 2
I (20) TASK 3: Value = 3
I (20) TASK 3: Value = 4

读取任务4

    void task_4(void *args)
    {
        uint32_t *RxValue = NULL;
    
        while (1)
        {
            vTaskDelay(pdMS_TO_TICKS(500));
            
            if (uxQueueMessagesWaiting(myQueue_2) > 0)
            {
                xQueueReceive(myQueue_2, &RxValue, (TickType_t)5);
                ESP_EARLY_LOGI("TASK 4", "Received from myQueue_2 = %d\n", *RxValue);
                ESP_EARLY_LOGI("TASK 4", "Data waiting in myQueue_2 : %d, Space available : %d", 
                uxQueueMessagesWaiting(myQueue_2), uxQueueSpacesAvailable(myQueue_2));
            }
        }
    }

任务4的产出

I (30) TASK 3: Data waiting in myQueue_2_1 : 5, Space available : 0
I (1300) TASK 4: Received from myQueue_2 = 5
I (1300) TASK 4: Data waiting in myQueue_2 : 4, Space available : 1
I (1800) TASK 4: Received from myQueue_2 = 5
I (1800) TASK 4: Data waiting in myQueue_2 : 3, Space available : 2
I (2300) TASK 4: Received from myQueue_2 = 5
I (2300) TASK 4: Data waiting in myQueue_2 : 2, Space available : 3
I (2800) TASK 4: Received from myQueue_2 = 5
I (2800) TASK 4: Data waiting in myQueue_2 : 1, Space available : 4
I (3300) TASK 4: Received from myQueue_2 = 5
I (3300) TASK 4: Data waiting in myQueue_2 : 0, Space available : 5

Answer 1

xQueueSend()发送指向值的指针，而不是值本身。它并不能复制被指的东西。如果指向的值在读取队列之前发生了更改，则读取器将看到新的值。

您的代码将相同的五个指针排队到变量count，而不是count值的五个副本。到task_4运行时，count等于5，所以task_4每次读取5。

正确的方法是每次传递一个不同的指针。您可以每次malloc()值并释放它：

for (count = 0; count < 5;count++)
    {
        uint32_t *value = (uint32_t *)malloc(sizeof(uint32_t));

        if (value) {
            *value = count;
            ESP_EARLY_LOGI("TASK 3", "Value = %d\n", *value);
            xQueueSend(myQueue_2, (void*)&Value, (TickType_t)0);
        } else {
            ESP_EARLY_LOGI("TASK 3", "malloc failed");
        }
    }

和

while (1)
    {
        vTaskDelay(pdMS_TO_TICKS(500));
        
        if (uxQueueMessagesWaiting(myQueue_2) > 0)
        {
            xQueueReceive(myQueue_2, &RxValue, (TickType_t)5);
            ESP_EARLY_LOGI("TASK 4", "Received from myQueue_2 = %d\n", *RxValue);
            ESP_EARLY_LOGI("TASK 4", "Data waiting in myQueue_2 : %d, Space available : %d", 
            uxQueueMessagesWaiting(myQueue_2), uxQueueSpacesAvailable(myQueue_2));
            free(RxValue);
        }
    }

第一个代码片段分配存储空间并每次对其进行排队，因此每个排队的项都将不同。非常重要的是，第二个代码片段释放存储空间，否则程序将泄漏内存并最终耗尽内存。

由于指针足够大以存储uint32_t，所以您也可以将计数器转换为void *，并在接收任务中将其转换回uint32_t。如果您想传递比指针更大的内容，则需要对其进行malloc()并初始化它，或者对要传递的对象执行自己的内存管理。