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Potential bug source in the code generated by CubeMX for components with shared clocks ?

SMarie
Senior

Hello,

I think I might have found a bug source in the CubeMX's generated code, when two components of a same driver share the same clock.

Here I'll use the example of the ADC 1 and 2 of the STM32H743GTV6 but I have see this case a few times elsewhere.

As the two ADC use the same clock, to avoid initiating it while the other is already using it, a check variable is used:

 

static uint32_t HAL_RCC_ADC12_CLK_ENABLED=0;

void HAL_ADC_MspInit(ADC_HandleTypeDef* adcHandle)
{

  GPIO_InitTypeDef GPIO_InitStruct = {0};
  if(adcHandle->Instance==ADC1)
  {
  /* USER CODE BEGIN ADC1_MspInit 0 */

  /* USER CODE END ADC1_MspInit 0 */
    /* ADC1 clock enable */
    HAL_RCC_ADC12_CLK_ENABLED++;
    if(HAL_RCC_ADC12_CLK_ENABLED==1){
      __HAL_RCC_ADC12_CLK_ENABLE();
    }
  
  /* Some code */

  }
  else if(adcHandle->Instance==ADC2)
  {
  /* USER CODE BEGIN ADC2_MspInit 0 */

  /* USER CODE END ADC2_MspInit 0 */
    /* ADC2 clock enable */
    HAL_RCC_ADC12_CLK_ENABLED++;
    if(HAL_RCC_ADC12_CLK_ENABLED==1){
      __HAL_RCC_ADC12_CLK_ENABLE();
    }

   /* Rest of the function */
  }
}

 

 

And the same thing is done is the void HAL_ADC_MspDeInit(ADC_HandleTypeDef* adcHandle) function:

 

    HAL_RCC_ADC12_CLK_ENABLED--;
    if(HAL_RCC_ADC12_CLK_ENABLED==0){
      __HAL_RCC_ADC12_CLK_DISABLE();
    }

 

But in the case where HAL_ADC_MspDeInit is called before HAL_ADC_MspInit, the HAL_RCC_ADC12_CLK_ENABLED (initiated to 0) will be 0xFFFF FFFF and so the init won't work.

I think something like that would be enough:

 

    if(HAL_RCC_ADC12_CLK_ENABLED!=0){
      HAL_RCC_ADC12_CLK_ENABLED--;
      if(HAL_RCC_ADC12_CLK_ENABLED==0){
        __HAL_RCC_ADC12_CLK_DISABLE();
      }
    }

 

Best regards,

Sulian

1 ACCEPTED SOLUTION

Accepted Solutions
Pavel A.
Evangelist III

Yes this is a known issue but a more complicated solution would be er... more complicated. So in the end, the developer is responsible to find a satisfying solution.

View solution in original post

1 REPLY 1
Pavel A.
Evangelist III

Yes this is a known issue but a more complicated solution would be er... more complicated. So in the end, the developer is responsible to find a satisfying solution.