Question

Basically what the question says, if I execute a MOVSD that isn’t 8-byte (or even 4-byte) aligned on various CPUs, what happens? Does it have a performance impact, can it segfault, etc.?

Answer 1

From Volume 1 of the Intel Spec, Section 4.1

4.1.1 Alignment of Words, Doublewords, Quadwords, and Double Quadwords

Words, doublewords, and quadwords do not need to be aligned in memory
on natural boundaries. The natural boundaries for words, double words,
and quadwords are even-numbered addresses, addresses evenly divisible
by four, and addresses evenly divisible by eight, respectively.
However, to improve the performance of programs, data structures
(especially stacks) should be aligned on natural boundaries when- ever
possible. The reason for this is that the processor requires two
memory accesses to make an unaligned memory access; aligned accesses
require only one memory access. A word or doubleword operand that
crosses a 4-byte boundary or a quadword operand that crosses an 8-byte
boundary is considered unaligned and requires two separate memory bus
cycles for access.

Under “normal” circumstances, an unalgined movsd will require 2 cycles to complete.

If you turn on alignment checking (in EFLAGS) then the CPU will raise an AC signal. It’s primarily meant as a way to help you detect unaligned accesses.