Is it possible to use analog watchdog without adc interrupt?

[SShir.2]

I am thinking of turning on ADC with DMA when analog watchdog detect certain level of signal. Then I would like to turn on ADC with only watchdog first and when LevelOutOfWindowCallback() is called ADC with DMA starts. I tried code below but it didn't work. It seems that watchdog never started or worked.

Please teach me how to do that.

Thank you.

int main(void){

configurations

HAL_ADC_Start(&hadc1);

 ADC1->CR1|=(1<<23);

 ADC1->CR1|=(1<<1);

 ADC1->SR &= ~(1 << 0);

 HAL_TIM_Base_Start(&htim2);

while(1){

}

}

void HAL_ADC_LevelOutOfWindowCallback(ADC_HandleTypeDef* hadc){

HAL_ADC_Start_DMA(&hadc1,(uint32_t*)DMA_buf,sizeof(DMA_buf)/sizeof(DMA_buf[0]));

}

[S.Ma]

If you use hal, use it too for watchdog interrupt too. And if you only need to trigger a range, DMA maybe only needed if you use ADC data as multichannel reading.

[SShir.2]

Thank you.

Just simply writing like below Both DMA and analog watchdog works. This is okay though, my ideal is to use ADC and watchdog separately. So I don't need any data which is out of threshold. So I would like to trigger to start DMA by LevelOutOfWindowCallback() . But when I only write HAL_ADC_Start_IT(&hadc1), LevelOutOfWindowCallback() is not called but when I write both  HAL_ADC_Start_IT and HAL_ADC_Start_DMA. Is it possible to use only watchdog first without using DMA ?

 HAL_ADC_Start_IT(&hadc1);

 HAL_ADC_Start_DMA(&hadc1,(uint32_t*)DMA_buf,sizeof(DMA_buf)/sizeof(DMA_buf[0]));