Can x86's MOV really be “free”? Why can't I reproduce this at all?

Question

I keep seeing people claim that the MOV instruction can be free in x86, because of register renaming.

For the life of me, I can\'t verify this in a single test cas

Answer 1

Here are two small tests that I believe conclusively show evidence for mov-elimination:

__loop1:
    add edx, 1
    add edx, 1
    add ecx, 1
    jnc __loop1

versus

__loop2:
    mov eax, edx
    add eax, 1
    mov edx, eax
    add edx, 1
    add ecx, 1
    jnc __loop2

If mov added a cycle to a dependency chain, it would be expected that the second version takes about 4 cycles per iteration. On my Haswell, both take about 2 cycles per iteration, which cannot happen without mov-elimination.