Agner`s CPU blog

mov ebp, 100
align 32
LL:
%rep 100 
                     ;  uops  Bytes
cmove eax,eax        ;    2      3
cmove ebx,ebx        ;    2      3
xchg  r8,r9          ;    3      3
nop7                 ;    1      7
nop7                 ;    1      7
nop8                 ;    1      8
nop                  ;    1      1
;                Total:  11     32
%endrep
dec ebp
jnz LL

This takes almost 4 clocks. When I add a nop after align 32 to change the alignment by one byte, it takes only 3 clocks. The explanation is this. Each µop cache line can take 6 µops. The first two instructions take one µop cache line. The xchg instruction cannot cross a cache line so it starts in a new cache line. The next three instructions go in the same line, and the last nop takes a third line. Then there is a 32-bytes boundary and we start a new cache line. In total we need 300 cache lines, and there are only 256 lines in the µop cache. The loop doesn't fit into the µop cache, so the decoders become the bottleneck. When the alignment is changed, the last nop goes together with the two cmove instructions in the next iteration, and we need only 200 cache lines. Now it fits into the µop cache and the speed goes up. The same can be obtained by lowering the repeat count.