谁能告诉我如何正确地将看门狗与DMA结合起来？

我想仅在其中一个通道上的电压超过特定限制后才在 ADC1 块的 2 个常规通道上启动 ADC 转换，因此我使用了看门狗功能。
触发看门狗后，我需要存储每个通道的 5 个结果值。这是我到目前为止的相关代码和平：

void HAL_ADC_LevelOutOfWindowCallback(ADC_HandleTypeDef *hadc)
{
HAL_ADC_Stop_IT(&hadc1); // stop watchdog interrupts
uint32_t adc_values[10] = {0};
HAL_ADC_Start_DMA(&hadc1, adc_values, 10); // read values from each channel after a watchdog has been triggered
HAL_ADC_Stop_DMA(&hadc1); // prevents continuous overwriting
uint32_t A4_val = 0;
uint32_t A5_val = 0;
A4_val = adc_values[1]; // values from channel A4
A5_val = adc_values[0]; // values from channel A5
printf("A4 channel value is %lun", A4_val);
printf("A5 channel value is %lun", A5_val);
A4_val = adc_values[3];
A5_val = adc_values[2];
printf("A4 channel value is %lun", A4_val);
printf("A5 channel value is %lun", A5_val);
A4_val = adc_values[5];
A5_val = adc_values[4];
printf("A4 channel value is %lun", A4_val);
printf("A5 channel value is %lun", A5_val);
A4_val = adc_values[7];
A5_val = adc_values[6];
printf("A4 channel value is %lun", A4_val);
printf("A5 channel value is %lun", A5_val);
A4_val = adc_values[9];
A5_val = adc_values[8];
printf("A4 channel value is %lun", A4_val);
printf("A5 channel value is %lun", A5_val);
printf("--------------n");
}