Nested structure padding - C - 64 bit - linux

Question

I have the following Nested structure in C. (64 bit)

    typedef struct {
        int a;
        int b;
        int c;
        struct {
            int ab;
            long bc;
        }
        int d;
    } Test;

I see that,
a = 4 bytes
b = 4 bytes
c = 4 bytes
padding1 = 4 bytes 
inner structure = 16 bytes ( 4 bytes for ab, 4 bytes padding, 8 bytes for bc)
d = 4 bytes
padding2 = 4 bytes

sizeof(Test) returns 40 bytes.

My questions:

1. padding1 -> why is this 4 bytes? Is this because the inner structure itself should be aligned ?. ( Also, Is it aligned with 8 byte (long) or 16 byte (size of inner) boundary.? )

2. padding2 -> Is this 4 byte padding because of max of alignment done inside the structure (which is 8) ??

Thanks,

Answer 1

1. padding1 -> why is this 4 bytes? Is this because the inner structure itself should be aligned ?. ( Also, Is it aligned with 8 byte (long) or 16 byte (size of inner) boundary.? )

It's because the inner struct should be 8-byte aligned, so that the long can reliably be 8-byte aligned.

1. padding2 -> Is this 4 byte padding because of max of alignment done inside the structure (which is eight) ??

It is there so that the size of the entire struct is a multiple of eight bytes, so that the inner struct can be suitably aligned on an eight-byte boundary.

In this particular case, the alignment requirements could be met with just four bytes of padding if an anonymous struct member could be treated differently from a free-standing struct, but 6.7.2.1

14 Each non-bit-field member of a structure or union object is aligned in an implementation-defined manner appropriate to its type.

forbids that. So to reduce the size of the struct, the programmer needs to rearrange it, move an odd number of int members past the inner struct (or make int ab; and long bc; direct members of Test without going through an anonymous struct).