Questions about C bitfields

Question

• Is bitfield a C concept or C++?

• Can it be used only within a structure? What are the other places we can use them?

• AFAIK, bitfields are special structure variables that occupy the memory only for specified no. of bits. It is useful in saving memory and nothing else. Am I correct?

I coded a small program to understand the usage of bitfields - But, I think it is not working as expected. I expect the size of the below structure to be 1+4+2 = 7 bytes (considering the size of unsigned int is 4 bytes on my machine), But to my surprise it turns out to be 12 bytes (4+4+4). Can anyone let me know why?

#include <stdio.h>

struct s{
unsigned int a:1;
unsigned int b;
unsigned int c:2;
};

int main()
{
  printf("sizeof struct s = %d bytes \n",sizeof(struct s));
  return 0;
}

OUTPUT:

sizeof struct s = 12 bytes 

Answer 1

Because a and c are not contiguous, they each reserve a full int's worth of memory space. If you move a and c together, the size of the struct becomes 8 bytes.

Moreover, you are telling the compiler that you want a to occupy only 1 bit, not 1 byte. So even though a and c next to each other should occupy only 3 bits total (still under a single byte), the combination of a and c still become word-aligned in memory on your 32-bit machine, hence occupying a full 4 bytes in addition to the int b.

Similarly, you would find that

struct s{
unsigned int b;
short s1;
short s2;
};

occupies 8 bytes, while

struct s{
short s1;
unsigned int b;
short s2;
};

occupies 12 bytes because in the latter case, the two shorts each sit in their own 32-bit alignment.