How to ensure a member is 4-byte aligned?

Question

In order to use OSAtomicDecrement (mac-specific atomic operation), I need to provide a 4-byte aligned SInt32.

Does this kind of cooking work ? Is there another way to deal with alignment issues ?

struct SomeClass {
  SomeClass() {
    member_  = &storage_ + ((4 - (&storage_ % 4)) % 4);
    *member_ = 0;
  }

  SInt32 *member_;

  struct {
    SInt32 a;
    SInt32 b;
  } storage_;
};

Answer 1

I know nothing about Mac programming but on minicomputers that I used to work on, pointers were always aligned on 4-byte (word) boundaries. IIRC, structs were too. Allocated memory always was.

Answer 2

If your compiler supports TR1 (or C++0x), you can use the std::aligned_storage template.

To allocate space for an object with size S and alignment A, you can allocate an object of type std::aligned_storage<S, A>::storage.

(The namespace may vary between compilers. I think TR1 doesn't specify which namespace the extensions must be placed in. On MSVC, the namespace std::tr1 is used)

Apart from this, 32-bit integers are already 4-byte aligned by the compiler (at least on the platforms where the natural alignment of 32-bit ints is 4 bytes)

Answer 3

ints are 4 byte aligned by default with any of the OS X compilers. All you need to do is not intentionally break that alignment (e.g. by doing improper pointer casts, marking your structure as packed, etc).

Answer 4

I would guess that any SInt32 is already aligned, even on a mac.

To clarify:

struct Foo {
    SInt32 member;
};

member is always aligned properly, unless you pack the structure and put member after a char.

Answer 5

If you're on a Mac, that means GCC. GCC can auto align variables for you:

  __attribute__((__aligned__(4))) int32_t member_;

Please note that this is not portable across compilers, as this is GCC specific.

Answer 6

If you want to force a nice alignment in a structure, you can use bit fields.

struct 
{
   Foo _hisFoo;
   unsigned int dummyAlignment : 0;
   Foo _myFoo;
}