On the 16 vs 15 byte-count padding front, the actual number you need to add to get an alignment of N is max(0,N-M) where M is the natural alignment of the memory allocator (and both are powers of 2).
Since the minimal memory alignment of any allocator is 1 byte, 15=max(0,16-1) is a conservative answer. However, if you know your memory allocator is going to give you 32-bit int aligned addresses (which is fairly common), you could have used 12 as a pad.
This isn't important for this example but it might be important on an embedded system with 12K of RAM where every single int saved counts.
The best way to implement it if you're actually going to try to save every byte possible is as a macro so you can feed it your native memory alignment. Again, this is probably only useful for embedded systems where you need to save every byte.
In the example below, on most systems, the value 1 is just fine for MEMORY_ALLOCATOR_NATIVE_ALIGNMENT, however for our theoretical embedded system with 32-bit aligned allocations, the following could save a tiny bit of precious memory:
#define MEMORY_ALLOCATOR_NATIVE_ALIGNMENT 4
#define ALIGN_PAD2(N,M) (((N)>(M)) ? ((N)-(M)) : 0)
#define ALIGN_PAD(N) ALIGN_PAD2((N), MEMORY_ALLOCATOR_NATIVE_ALIGNMENT)