I have a structure that looks like this:
struct vdata {
static_assert(sizeof(uint8_t *) == 8L, "size of pointer must be 8");
union union_data {
uint8_t * A; // 8 bytes
uint8_t B[12]; // 12 bytes
} u;
int16_t C; // 2 bytes
int16_t D; // 2 bytes
};
I would like to make this 16 bytes, but GCC is telling me it is 24, as the union is padding to 16 bytes.
I would like to put vdata into a large std::vector
. From my understanding, there should be no issue with alignment if this were 16 bytes, since the pointer would always be 8 byte aligned.
I understand that I can force this to be packed using __attribute__((__packed__))
in GCC. But I would like to know if there is a portable and standard compliant way to get this to be 16 bytes?
Edit: Ideas
Idea 1: split up the B array.
struct vdata {
union union_data {
uint8_t * A; // 8 bytes
uint8_t B[8]; // 8 bytes
} u;
uint8_t B2[4]; // 4 bytes
int16_t C; // 2 bytes
int16_t D; // 2 bytes
};
Could B2 elements be reliably accessed from a pointer of B? Is that defined behavior?
Idea 2: store pointer as byte array and memcpy as necessary (@Eljay)
struct vdata {
union union_data {
std::byte A[sizeof(uint8_t*)]; // 8 bytes
uint8_t B[12]; // 12 bytes
} u;
int16_t C; // 2 bytes
int16_t D; // 2 bytes
};
Would there be a performance penalty for accessing the pointer, or would it be optimized out? (Assuming GCC x86).