64 bit architecture like x86-64 have word size of 64bits. In this case, if a memory access crosses over the word boundary, then it will require double the time to access data. So alignment is required. - This is what I know. Correct me if I am wrong.
Now, GCC uses 16 byte alignment (msvc atleast uses 8 byte alignment) for long double whose non-padding size is 10 bytes. But anyways, with 8 byte alignment it requires 2 read cycles and it is the same case with 16 byte alignment. So why stricter 16 byte alignment? What is the purpose of alignment other than that I mentioned above?
Also, in fact, since the non-padding part of long double (the 80-bit x87 extended FP) is 10 bytes, actually 4 byte alignment is sufficient for that. In this case also, it can read data within 2 read cycles (either 4-6 or 8-2). So, also explain where this assumption has gone wrong.
(The actual sizeof(long double) is 12 in the i386 System V ABI, 16 in x86-64 System V. Multiples of their respective alignof() of 4 and 16)
fld m80does decode into a 2-byte and an 8-byte load, so 8 or 16 byte alignment are both sufficient to avoid cache-line splits in either of the halves. But only if the size is 16 bytes, so you might as well make the align match the size for SSE copying.long double, but it does not explain why or what the impact is. My quick experiments showed no impact of (mis)alignment, only of crossing cache line boundaries, as expected.