Memory layout of union of different sized member?
typedef union epoll_data
{
void *ptr;
int fd;
__uint32_t u32;
__uint64_t u64;
} epoll_data_t;
Here int
and __uint32_t
are 4 bytes,while the others are 8 bytes.
When we set fd
to an int
,does it lie on the first 4 bytes or 开发者_如何学Gothe last 4 bytes,or does it depend on endianness?
Some reason is appreciated.
It lies on the first 4 bytes. From the C99 standard §6.7.2.1/14 (§6.7.2.1/16 in C11 and C18):
The size of a union is sufficient to contain the largest of its members. The value of at most one of the members can be stored in a union object at any time. A pointer to a union object, suitably converted, points to each of its members (or if a member is a bit-field, then to the unit in which it resides), and vice versa.
This implies that the address of all members of a union is the same.
This really depends on ELF-ABI for that platform. See examples and figures give under section 3.1 in http://www.sco.com/developers/devspecs/abi386-4.pdf It shows that it need not start at low address, if there is padding due to alignment constraints.
精彩评论