3 ms·

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.

• 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.