Copy unaligned XMVECTOR and use the copy?

Answering the question after the edit:

If XMVECTOR was allocated on the stack then the compiler will automatically force it to be 16-byte aligned.

The only case that you have an unaligned XMVECTOR is when it's allocated on the heap, but should't be used at all, because it will result in a seg-fault.

On the other hand if you are accessing the XMVECTOR as a memory chunk using its address, in other words without doing any computation it won't be copied to the SSE registers. You can copy it just like any other POD. The problem isn't accessing the memory, the problem is about using it, so copying it is OK as far as you don't do any math operations that use SSE registers.

If you are only building for x64 native, this code is perfectly fine since it is always 16-byte aligned.

class A { public: XMVECTOR vect; }; void fun(void) { A* a = new A; XMVECTOR localVect = a.vect; } 

To make it portable to x86 or ARM (32-bit), you have one of three choices:

1. Create an overloaded new for the class A that uses __aligned_malloc(,16);
2. Create a global new overload that does this (which is not recommended)
3. Just use XMFLOAT4 or it's kin and explicitly use XMLoadFloat4/XMStoreFloat4

The following code results in the identical compiled code as the block above except that one movaps is now a movups:

class A { public: XMFLOAT4 vect; }; void fun(void) { A* a = new A; XMVECTOR localVect = XMLoadFloat4( &a.vect ); } 

As such, there's no point in messing around with aligned allocations unless it's trivial to do for some other reason (such as DirectX Tool Kit's use of the pImpl idiom which means we might as well align the internal class allocations to make it convenient).

If your complaint is just the extra work it takes to use XMVECTOR correctly with respect to the alignment, you could take a look at using SimpleMath from the DirectX Tool Kit.