64位计算机上的struct alignment

Question

I have the following struct on a 64-bit Linux machine.

struct __wait_queue_head {
          spinlock_t lock;
          struct list_head task_list;
  };

where

typedef struct {
          raw_spinlock_t raw_lock;
  } spinlock_t;

and

struct list_head {
          struct list_head *next, *prev;
  };

raw_spinlock_t is defined as:

typedef struct {
          volatile unsigned int slock;
  } raw_spinlock_t;

Now I want to understand the alignment of the struct __wait_queue_head on a 64-bit Linux machine following the LP64 standard. From what I know, since the first field of this struct ie.

spinlock_t lock

is an unsigned int, which occupies 4 bytes on a 64-bit machine, this struct should begin at a 4-byte aligned address. However, I have seen that is not the case on a real system. Instead, the struct begins at an 8 byte aligned address, although the alignment requirement of the first field would have been met by a 4 byte aligned address. Basically, what governs alignment of a struct? Please note that I am clear about the padding concept of fields within a struct. The alignment requirement of a struct itself is what I am finding confusing.

Answer 1

The alignment requirement of a struct is the biggest alignment requirement of any of its members. In this case, since struct list_head contains pointers, the alignment of struct list_head is 8 bytes. And since struct __wait_queue_head contains a struct list_head , its alignment is 8 bytes as well. This is required because if the struct had a looser alignment requirement, then the struct padding wouldn't be enough to guarantee that the members would be properly aligned.