Store an address and a bool in one word for lock-f

Question

I am reading some paper for lock-free doubly linked list. In these papers, they store an address to next and prev node and a flag in one word(int).

Is it because in 32-bit architectures all addresses are aligned in 4 byte boundaries so all address are multiple of 4?

and if the reason is what i say is this code ok?

const int dMask = 1;
const int pMask = ~dMask;

int store(void* pPointer, bool pDel)
{
    return reinterpret_cast<int>(pPointer) | (int)pDel;
}

void load(int pData, void** pPointer, bool* pDel)
{
    *pPointer = reinterpret_cast<void*>(pData & pMask);
    *pDel = pData & dMask;
}

And another question: Do in other platforms such as Android mobile devices, above idea is correct?

Answer 1

Addresses in most 32-bit architectures are not stored in 4-byte boundaries, but 1-byte. They are read from memory at 4-byte (the typical word size) increments. Without seeing the code for this doubly linked list, it sounds like they are enforcing some rules for how the container will store data.

Answer 2

You're more or less correct. It's a common space optimization in very low level code. It is not portable. (You could make it slightly more portable by using intptr_t instead of int.)

Also, of course, the alignment only holds for pointers to more complex types; a char* won't necessarily be aligned. (The only times I've seen this used is in the implementation of memory management, where all of the blocks involved are required to be aligned sufficiently for any type.)

Finally, I'm not sure what the authors' of the paper are trying to do, but the code you post cannot be used in a multithreaded environment, at least on modern machines. To ensure that a modification in one thread is seen in another thread, you need to use atomic types, or some sort of fence or membar.