FreeRTOS的编译器优化问题

Question

我试图用一个简单的例子来解释我的问题

typedef function<bool()> TaskCallback;

class Task
{
public:
    Task(TaskCallback task_callback) : task_callback(task_callback)
    {
        long_string_test = "This is a long string 0123456789ABCDEF 0123456789ABCDEF 0123456789ABCDEF";

        xTaskCreate(Task::RunTask, "task_name", 2560, this, 3, &task_handle);
    }

    ~Task()
    {
        while(1); //Breakpoint: The destructor is never called
    }

private:
    static void RunTask(void* params)
    {
        Task* _this = static_cast<Task*>(params);

        _this->task_callback(); //The program crashes here because task_callback doesn't exist
    }

    string long_string_test;

    TaskCallback task_callback;

    TaskHandle_t task_handle;
};

main.cpp

static bool Init_task() { }

void main()
{
    Task task(Init_task);

    vTaskStartScheduler();

    //We should never get here as control is now taken by the FreeRTOS scheduler
    while(1);
}

如果通过long_string_test函数中的调试器检查字符串RunTask的值，我会发现它具有一个奇怪的值，就好像该字符串已被破坏。但是从未调用过Task类的析构函数。

如果我在程序下面更改“ main.cpp”，则该程序可以正常工作，我认为编译器会进行某种优化：

static bool Init_task() { }

Task task(Init_task);

void main()
{
    vTaskStartScheduler();

    //We should never get here as control is now taken by the FreeRTOS scheduler
    while(1);
}

ps.s。显然，编译器优化已禁用

Answer 1

Answer 2

该实现可能假定任何线程最终都会执行以下操作之一：

terminate，

• 调用库I / O函数，
• 通过易失的glvalue执行访问，或
• 执行同步操作或原子操作。

[注意：这旨在允许编译器转换，例如删除空循环，即使无法证明终止也是如此。 —尾注]

和

~Task()
{
    while(1); //Breakpoint: The destructor is never called
}

都不做，因此编译器是免费的，而忽略它。

Answer 3

int* test;

static void RunTask(void* params)
{
    Print(*test);  //The "test" pointer has a random value
}

void main()
{
    int temp = 9999;

    test = &temp;

    xTaskCreate(RunTask, "task_name", 2560, NULL, 3, NULL);

    vTaskStartScheduler(); //It seems that FreeRTOS clears the main() stack

    //We should never get here as control is now taken by the FreeRTOS scheduler
    while(1);
}