size of a structure containing bit fields

Question

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Why isn't sizeof for a struct equal to the sum of sizeof of each member?

I was trying to understand the concept of bit fields. But I am not able to find why the size of the following structure in CASE III is coming out as 8 bytes.

CASE I:

struct B    
{
    unsigned char c;  // +8 bits
} b;

sizeof(b); // Output: 1 (because unsigned char takes 1 byte on my system)

CASE II:

struct B
{
    unsigned b: 1;
} b;

 sizeof(b); // Output: 4 (because unsigned takes 4 bytes on my system)

CASE III:

struct B
{
    unsigned char c;  // +8 bits
    unsigned b: 1;    // +1 bit
} b;

sizeof(b); // Output: 8 

I don't understand why the output for case III comes as 8. I was expecting 1(char) + 4(unsigned) = 5.

Answer 1

You can check the layout of the struct by using offsetof, but it will be something along the lines of:

struct B
{
    unsigned char c;  // +8 bits
    unsigned char pad[3]; //padding
    unsigned int bint; //your b:1 will be the first byte of this one
} b;

Now, it is obvious that (in a 32-bit arch.) the sizeof(b) will be 8, isn't it?

The question is, why 3 bytes of padding, and not more or less?

The answer is that the offset of a field into a struct has the same alignment requirements as the type of the field itself. In your architecture, integers are 4-byte-aligned, so offsetof(b, bint) must be multiple of 4. It cannot be 0, because there is the c before, so it will be 4. If field bint starts at offset 4 and is 4 bytes long, then the size of the struct is 8.

Another way to look at it is that the alignment requirement of a struct is the biggest of any of its fields, so this B will be 4-byte-aligned (as it is your bit field). But the size of a type must be a multiple of the alignment, 4 is not enough, so it will be 8.