STM32时钟系统初始化程序遇到的问题

Question

STM32时钟系统初始化程序

static void SetSysClockTo72(void)
{
  __IO uint32_t StartUpCounter = 0, HSEStatus = 0;
  
  /* SYSCLK, HCLK, PCLK2 and PCLK1 configuration ---------------------------*/    
  /* Enable HSE */    
  RCC->CR |= ((uint32_t)RCC_CR_HSEON);
 
  /* Wait till HSE is ready and if Time out is reached exit */
  do
  {
    HSEStatus = RCC->CR & RCC_CR_HSERDY;
    StartUpCounter++;  
  } while((HSEStatus == 0) && (StartUpCounter != HSE_STARTUP_TIMEOUT));

  if ((RCC->CR & RCC_CR_HSERDY) != RESET)
  {
    HSEStatus = (uint32_t)0x01;
  }
  else
  {
    HSEStatus = (uint32_t)0x00;
  }  

  if (HSEStatus == (uint32_t)0x01)
  {
    /* Enable Prefetch Buffer */
    FLASH->ACR |= FLASH_ACR_PRFTBE;

    /* Flash 2 wait state */
    FLASH->ACR &= (uint32_t)((uint32_t)~FLASH_ACR_LATENCY);
    FLASH->ACR |= (uint32_t)FLASH_ACR_LATENCY_2;    

 
    /* HCLK = SYSCLK */
    RCC->CFGR |= (uint32_t)RCC_CFGR_HPRE_DIV1;
      
    /* PCLK2 = HCLK */
    RCC->CFGR |= (uint32_t)RCC_CFGR_PPRE2_DIV1;
    
    /* PCLK1 = HCLK/2 */
    RCC->CFGR |= (uint32_t)RCC_CFGR_PPRE1_DIV2;

#ifdef STM32F10X_CL
    /* Configure PLLs ------------------------------------------------------*/
    /* PLL2 configuration: PLL2CLK = (HSE / 5) * 8 = 40 MHz */
    /* PREDIV1 configuration: PREDIV1CLK = PLL2 / 5 = 8 MHz */
        
    RCC->CFGR2 &= (uint32_t)~(RCC_CFGR2_PREDIV2 | RCC_CFGR2_PLL2MUL |
                              RCC_CFGR2_PREDIV1 | RCC_CFGR2_PREDIV1SRC);
    RCC->CFGR2 |= (uint32_t)(RCC_CFGR2_PREDIV2_DIV5 | RCC_CFGR2_PLL2MUL8 |
                             RCC_CFGR2_PREDIV1SRC_PLL2 | RCC_CFGR2_PREDIV1_DIV5);
  
    /* Enable PLL2 */
    RCC->CR |= RCC_CR_PLL2ON;
    /* Wait till PLL2 is ready */
    while((RCC->CR & RCC_CR_PLL2RDY) == 0)
    {
    }
    
   
    /* PLL configuration: PLLCLK = PREDIV1 * 9 = 72 MHz */ 
    RCC->CFGR &= (uint32_t)~(RCC_CFGR_PLLXTPRE | RCC_CFGR_PLLSRC | RCC_CFGR_PLLMULL);
    RCC->CFGR |= (uint32_t)(RCC_CFGR_PLLXTPRE_PREDIV1 | RCC_CFGR_PLLSRC_PREDIV1 | 
                            RCC_CFGR_PLLMULL9); 
#else    
    /*  PLL configuration: PLLCLK = HSE * 9 = 72 MHz */
    RCC->CFGR &= (uint32_t)((uint32_t)~(RCC_CFGR_PLLSRC | RCC_CFGR_PLLXTPRE |
                                        RCC_CFGR_PLLMULL));
    RCC->CFGR |= (uint32_t)(RCC_CFGR_PLLSRC_HSE | RCC_CFGR_PLLMULL9);
#endif /* STM32F10X_CL */

    /* Enable PLL */
    RCC->CR |= RCC_CR_PLLON;

    /* Wait till PLL is ready */
    while((RCC->CR & RCC_CR_PLLRDY) == 0)
    {
    }
    
    /* Select PLL as system clock source */
    RCC->CFGR &= (uint32_t)((uint32_t)~(RCC_CFGR_SW));
    RCC->CFGR |= (uint32_t)RCC_CFGR_SW_PLL;    

    /* Wait till PLL is used as system clock source */
    while ((RCC->CFGR & (uint32_t)RCC_CFGR_SWS) != (uint32_t)0x08)
    {
    }
  }
  else
  { /* If HSE fails to start-up, the application will have wrong clock 
         configuration. User can add here some code to deal with this error */
  }
}

程序里面的
do
{
HSEStatus = RCC->CR & RCC_CR_HSERDY;
StartUpCounter++;
} while((HSEStatus == 0) && (StartUpCounter != HSE_STARTUP_TIMEOUT));
和
while((RCC->CR & RCC_CR_PLLRDY) == 0)
{
}

这两段代码都是检测时钟源是否就绪，对RCC->CR使能后为0x01000000，RCC_CR_PLLRDY就绪后为0x02000000，&运算后值不就等于零了吗，那这样子不就进入死循环了吗。