5 ms·

When Linus made that comment cmov was like a 6 cycle latency. For the last decade it has been 1 cycle, and I don't think there is any scenario where cmov is now

When Linus made that comment cmov was like a 6 cycle latency. For the last decade it has been 1 cycle, and I don't think there is any scenario where cmov is now slower than a branch.

haberman 2y ago
The problem is not the 1 cycle latency, but the data dependency on both values. A correctly-predicted branch cuts the dependency on the value that is not used. I've definitely measured scenarios where cmov/branchless is slower than a branch for a given algorithm. Especially if the branchless version is doing a bit more work to avoid the branch.
- unnah 2y ago
  Good point. It makes me wonder if modern out-of-order processors can skip performing unused computations altogether, if their result registers are overwritten by other data later (in program order).
  - ants_a 2y ago
    No, and it feels unlikely that they will either.
  - gpderetta 2y ago
    CPUs could in theory speculate CMOV, reintroducing prediction when predictable. But after Spectre, IIRC Intel now guarantees that CMOV is never speculated.
- gpderetta 2y ago
  It is both though. At 6+ cycles there are only a few places where CMOV is a win. At 1 cycle you can be more liberal with its use and the lack of dependency breaking is a tradeoff.
- jnordwick 2y ago
  cmov turns a control dependency into a data dependency. It is still there just in a different way. (its like the Fourier Transform of instructions - lolol). I haven't seen cmov lose to a conditional in a few years (on very recent Intel hardware). Maybe there might be some circumstances where in creates a loop carried dependency that wrecks the optimizer, but I haven't seen one in a minute. To be fair - cmov versions often involve some extra compution to get things in a form where cmov can be used, and I often just opt for the well predicted conditional since those couple extra instructions to set up the value to move can be avoided.
mgaunard 2y ago
a cmov-based approach is necessarily slower than a branch-based approach that was correctly predicted, since cmov requires computing both branches then selecting the result at the end.
clausecker 2y ago
Are you sure? I recall cmov always having single cycle latency.
- BoardsOfCanada 2y ago
  I know that it was decoded into 2 micro-ops and thus had to be decoded by the wide decoder, so perhaps 2 cycles?
  - clausecker 2y ago
    That could be. Agner's tables seem to confirm that.
    - jnordwick 2y ago
      Per Fog's tables, Ice Lake shows a reg-to-reg cmov decoding to 1 uop with a latency of 1 and throughput of 0.5. *mont all have latency 2. P4 had a latency of 6. AMD's K10 latency 4.

When Linus made that comment cmov was like a 6 cycle latency. For the last decade it has been 1 cycle, and I don't think there is any scenario where cmov is now slower than a branch.

haberman 2y ago

The problem is not the 1 cycle latency, but the data dependency on both values. A correctly-predicted branch cuts the dependency on the value that is not used. I've definitely measured scenarios where cmov/branchless is slower than a branch for a given algorithm. Especially if the branchless version is doing a bit more work to avoid the branch.

unnah 2y ago
Good point. It makes me wonder if modern out-of-order processors can skip performing unused computations altogether, if their result registers are overwritten by other data later (in program order).
- ants_a 2y ago
  No, and it feels unlikely that they will either.
- gpderetta 2y ago
  CPUs could in theory speculate CMOV, reintroducing prediction when predictable. But after Spectre, IIRC Intel now guarantees that CMOV is never speculated.
gpderetta 2y ago
It is both though. At 6+ cycles there are only a few places where CMOV is a win. At 1 cycle you can be more liberal with its use and the lack of dependency breaking is a tradeoff.
jnordwick 2y ago
cmov turns a control dependency into a data dependency. It is still there just in a different way. (its like the Fourier Transform of instructions - lolol). I haven't seen cmov lose to a conditional in a few years (on very recent Intel hardware). Maybe there might be some circumstances where in creates a loop carried dependency that wrecks the optimizer, but I haven't seen one in a minute. To be fair - cmov versions often involve some extra compution to get things in a form where cmov can be used, and I often just opt for the well predicted conditional since those couple extra instructions to set up the value to move can be avoided.

mgaunard 2y ago

a cmov-based approach is necessarily slower than a branch-based approach that was correctly predicted, since cmov requires computing both branches then selecting the result at the end.

clausecker 2y ago

Are you sure? I recall cmov always having single cycle latency.

BoardsOfCanada 2y ago
I know that it was decoded into 2 micro-ops and thus had to be decoded by the wide decoder, so perhaps 2 cycles?
- clausecker 2y ago
  That could be. Agner's tables seem to confirm that.
  - jnordwick 2y ago
    Per Fog's tables, Ice Lake shows a reg-to-reg cmov decoding to 1 uop with a latency of 1 and throughput of 0.5. *mont all have latency 2. P4 had a latency of 6. AMD's K10 latency 4.