Is an extra move somehow faster when doing division-by-multiplication?

This very much looks like a missed optimization by both gcc and clang; no benefit to that extra mov.

If it's not already reported, GCC and LLVM both accept missed-optimization bug reports: and For GCC there's even a bug tag "missed-optimization".

Wasted mov instructions are unfortunately not rare, especially when looking at tiny functions where the input / output regs are nailed down the calling convention, not up to the register allocator. The do still happen in loops sometimes, like doing a bunch of extra work each iteration so everything is in the right places for the code that runs once after a loop. /facepalm.

Zero-latency mov (mov-elimination) helps reduce the cost of such missed optimizations (and cases where mov isn't avoidable), but it still takes a front-end uop so it's pretty much strictly worse. (Except by chance where it helps alignment of something later, but if that's the reason then a nop would have been as good).

And it takes up space in the ROB, reducing how far ahead out-of-order exec can see past a cache miss or other stall. mov is never truly free, only the execution-unit and latency part is eliminated - Can x86's MOV really be "free"? Why can't I reproduce this at all?

My total guess about compiler internals:

Probably gcc/clang's internal machinery need to learn that this division pattern is commutative and can take the input value in some other register and put the constant in RAX.

In a loop they'd want the constant in some other register so they could reuse it, but hopefully the compiler could still figure that out for cases where it's useful.

Visual Studio 2015 generates the code you expected, rcx = input dividend:

        mov     rax, 2635249153387078803
        mul     rcx
        sub     rcx, rdx
        shr     rcx, 1
        lea     rax, QWORD PTR [rdx+rcx]
        shr     rax, 2

A divisor of 7 needs a 65 bit multiplier to get the proper accuracy.

floor((2^(64+ceil(log2(7))))/7)+1 = floor((2^67)/7)+1 = 21081993227096630419

Removing the most significant bit, 2^64, results in 21081993227096630419 - 2^64 = 2635249153387078803, which is the multiplier actually used in the code.

The generated code compensates for the missing 2^64 bit, which is explained in figure 4.1 and equation 4.5 in this pdf file:

Further explanation can be seen in this prior answer:

Why does GCC use multiplication by a strange number in implementing integer division?

If the 65 bit multiplier has a trailing 0 bit, then it can be shifted right 1 bit to result in a 64 bit multiplier, reducing the number of instructions. For example if dividing by 5:

floor((2^(64+ceil(log2(5))))/5)+1 = floor((2^67)/5)+1 = 29514790517935282586
29514790517935282586 >> 1 = 14757395258967641293

        mov     rax, -3689348814741910323 ; == 14757395258967641293 ==  0cccccccccccccccdH
        mul     rcx
        shr     rdx, 2
        mov     rax, rdx