问题:

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 the last 4 bytes,or does it depend on endianness?

Some reason is appreciated.

回答1:

It lies on the first 4 bytes. From the C99 standard §6.7.2.1/14:

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.

回答2:

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.