Is there a way to access individual bits with a union?

Question

I am writing a C program. I want a variable that I can access as a char but I can also access the specific bits of. I was thinking I could use a union like this...

typedef union 
{
    unsigned char status;
    bit bits[8];
}DeviceStatus;

but the compiler doesn't like this. Apparently you can't use bits in a structure. So what can I do instead?

Answer 1

Sure, but you actually want to use a struct to define the bits like this

typedef union
{
  struct
  {
    unsigned char bit1 : 1;
    unsigned char bit2 : 1;
    unsigned char bit3 : 1;
    unsigned char bit4 : 1;
    unsigned char bit5 : 1;
    unsigned char bit6 : 1;
    unsigned char bit7 : 1;
    unsigned char bit8 : 1;
  }u;
  unsigned char status;
}DeviceStatus;

Then you can access for DeviceStatus ds; you can access ds.u.bit1. Also, some compilers will actually allow you to have anonymous structures within a union, such that you can just access ds.bit1 if you ommit the u from the typedef.

Answer 2

You have a couple of possibilities. One would be to just use Boolean math to get at the bits:

int bit0 = 1;
int bit1 = 2;
int bit2 = 4;
int bit3 = 8;
int bit4 = 16;
int bit5 = 32;
int bit6 = 64;
int bit7 = 128;

if (status & bit1)
    // whatever...

Another is to use bitfields:

struct bits { 
   unsigned bit0 : 1;
   unsigned bit1 : 1;
   unsigned bit2 : 1;
// ...
};

typedef union {
    unsigned char status;
    struct bits bits;
} status_byte;

some_status_byte.status = whatever;
if (status_byte.bits.bit2)
    // whatever...

The first is (at least arguably) more portable, but when you're dealing with status bits, chances are that the code isn't even slightly portable anyway, so you may not care much about that...