Question

* The ADC has been set up to sequence both analog inputs. * The interrupt is at the end of the sequence after both conversions. * Without DMA, only the last value is read. volatile static uint16_t ai; volatile static uint8_t got_ai - @; #ifdef DO QUESTION_2 void HAL_ADC_ConvCpltcallback (ADC_HandleTypeDef* hadc) { { if (hadc == &hadci) ai - (uint16_t) HAL_ADC_GetValue (&hadc1); got_ai - 1; } } Fendif void ADC_Interrupt (void) { printf ("ADC_Interrupt(): \n"); HAL_Status TypeDef status; uint32_t i; while (1) // printf("-> Start conversion... \n"); got_ai - ; if (status = HAL_ADC_Start_IT (&hadc1)) != HAL_OK) printf ("-> ERROR: HAL_ADC_Start_IT() call failed (status = %d)\n", status); return; { } // printf("-> Wait for conversion to finish... \n"); while (got_ai -- ) // Critical section protection not included { } // Print the result as decimal and bar graph printf ("Got %4d: ", ai); for (i = 0; i < 60; i++) { if (i <= ai * 60 / 4896) printf ("*"); { } { } else printf (" "); } printf ("|\r"); // Loop delay HAL_Delay (100); }

There is still a busy wait in the main loop, so what is the benefit of interrupts?

Answer 1

Solution: Busy waiting is a state of the processor or the executing unit in which the process repeatedly checks for a condition to be true while waiting in the queue. An interrupt is nothing but an instruction or a signal that is sent by the processor to change the state of the process so that it can be executed efficiently and effectively by the processor.

Now, coming to this question, the interrupts are meant to change the state the processes that are present in the queue busy waiting but in this case, they will anyway in a busy waiting state due to an infinite loop in the main function. Due to this loop, it will keep on waiting for an infinite amount of time in the busy waiting state.

Here's the solution to your question, please provide it a 100% rating. Thanks for asking and happy learning!!