关于STM32库中超时函数tick溢出的分析

STM32的官方库文件stm32f1xx_hal.c部分摘录如下

__IO uint32_t uwTick;

/**
  * @brief This function is called to increment  a global variable "uwTick"
  *        used as application time base.
  * @note In the default implementation, this variable is incremented each 1ms
  *       in SysTick ISR.
  * @note This function is declared as __weak to be overwritten in case of other
  *      implementations in user file.
  * @retval None
  */
__weak void HAL_IncTick(void)
{
  uwTick += uwTickFreq;
}

/**
  * @brief Provides a tick value in millisecond.
  * @note  This function is declared as __weak to be overwritten in case of other
  *       implementations in user file.
  * @retval tick value
  */
__weak uint32_t HAL_GetTick(void)
{
  return uwTick;
}

/**
  * @brief This function provides minimum delay (in milliseconds) based
  *        on variable incremented.
  * @note In the default implementation , SysTick timer is the source of time base.
  *       It is used to generate interrupts at regular time intervals where uwTick
  *       is incremented.
  * @note This function is declared as __weak to be overwritten in case of other
  *       implementations in user file.
  * @param Delay specifies the delay time length, in milliseconds.
  * @retval None
  */
__weak void HAL_Delay(uint32_t Delay)
{
  uint32_t tickstart = HAL_GetTick();
  uint32_t wait = Delay;

  /* Add a freq to guarantee minimum wait */
  if (wait < HAL_MAX_DELAY)
  {
    wait += (uint32_t)(uwTickFreq);
  }

  while ((HAL_GetTick() - tickstart) < wait)
  {
  }
}

如果工程中开启了systick功能，默认是1ms一个中断，每次 uwTick增加1，上述代码的uwTickFreq就等于1，具体可以查看底层的定义。

那么在HAL_Delay中，使用就是uwTick来计算延迟，但是uwTick是一个uint32_t的变量，会不会因为溢出，而导致延迟时间变化呢？

举例如下：

1）要延迟10ms，即Delay = 10（此处，wait有个初始判断+1，可以不用关注，延迟11ms，Delay = 11）

2）第一种情况，不存在uwTick溢出，调用HAL_Delay时，tickstart获取的tick假设是30000，然后当uwTick到30011时延迟结束

3）第二种情况，存在uwTick溢出，调用HAL_Delay时，tickstart获取的tick假设是65530，但是wait是11，那么当uwtick=65535+1，就会溢出uwtick = 0，此时计算uwtick-tickstart=-65530：二进制表示0000 0000 0000 0110,无符号十进制6，那么当uwtick=5（0000 0000 0000 0101）时，延时还是11ms（0000 0000 0000 1011），所以无符号的uwtick溢出并不会导致延迟错乱，此处差值不存在负数，要考虑好无符号数的计算方法

这个问题的主要依据是使用了计算机的原码，反码，补码的知识。可以参考如下链接,写的非常好

https://www.cnblogs.com/zhangziqiu/archive/2011/03/30/ComputerCode.html