Size of structure with bit fields

Question

Here I have a code snippet.

#include <stdio.h>
int main()
{
    struct value
    {
        int bit1 : 1;
        int bit2 : 4;
        int bit3 : 4;
    } bit;

    printf("%d",sizeof(bit));

    return 0;
}

I'm getting the output as 4 (32 bit compiler). Can anyone explain me how? Why is it not 1+ 4 + 4 = 9? I've never worked with bit fields before so would love some help. Thank you. :)

Answer 1

When you tell the C compiler this:

int bit1 : 1

It interprets it as, and allocates to it, an integer; but refers to it's first bit as bit1.

So if we consider your code:

struct value
{
    int bit1 : 1;
    int bit2 : 4;
    int bit3 : 4;
} bit;

What you are telling the compiler is this: Take necessary number of the ints, and refer to the chunks bit 1 as bit1, then refer to bits 2 - 5 as bit2, and then refer to bits 6 - 9 as bit3.

Since the complete number of bits required are 9, and an int is 32 bits (in your computer's architecture), memory space of only 1 int is required. Thus you get the size as 4 (bytes).

Instead, if you were to define the struct using chars, since char is 8 bits, the compiler would allocate the memory space of two chars for each struct value. And you will get 2 (bytes) as your output.

Answer 2

Because C requests to pack the bits in the same unit (here one signed int / unsigned int):

(C99, 6.7.2.1p10) "If enough space remains, a bit-field that immediately follows another bit-field in a structure shall be packed into adjacent bits of the same unit"

Answer 3

The processor just likes chucking around 32 bits in one go - not 9, 34 etc.

It just rounds it up to what the processor likes. (Keep the worker happy)