Decode ADC Result Using Union Bitfields

Code

#include <stdio.h>
#include <stdint.h>

#define ADC_VALUE(reg)      (reg & 0x0FFF)
#define CHANNEL_VALUE(reg)  ((reg & 0xF000) >> 12U)


typedef union {
    struct {
        uint16_t adc_value : 12;
        uint16_t channel   : 4;
    };
    uint16_t adc_reg;
} ADC_Result;

int main() {
    uint16_t reg;
    scanf("%hx", &reg);

    // Fill union and print channel and adc_value
    uint16_t adc_val = ADC_VALUE(reg);
    uint8_t channel = CHANNEL_VALUE(reg);

    printf("Channel: %d\n", channel);
    printf("ADC Value: %d\n", adc_val);

    return 0;
}

Solving Approach

Bitfields

Input: 0x10FF
Output:
Channel: 1  
ADC Value: 255

0x10FF -> 0b 0001 0000 1111 1111
for channel the mask -> 0xF000
shift to the right by 12 to get the channel
for adc-val the mask -> 0x0FFF



Plan:
- create a local object from the union
- create a mask for the adc value
- create a mask for the channel value
- print the channel and the adc value