Conditional moves vs. branching in unrolled loops

[Paul Sandoz]

> On 6 Jan 2016, at 15:45, Vitaly Davidovich <vitalyd at gmail.com> wrote:
> 
> Ok. The generated code for an unrolled loop firsts load array elements into registers before performing the cmovs.
> 
> Yes, but the cmov cannot proceed until that load retires.  If you had a normal branch, speculation can continue past the branch and put more instructions into the pipeline barring other hazards/dependencies.  By "available in registers" I meant a cmov executed against 2 values in registers that are already available (i.e. the loads which put the values into registers have already completed, or the registers were set with immediates, etc).
> 
> Basically, if the cost of branch misprediction is higher than waiting for both inputs to cmov to be available, then cmov is better.  For very predictable branches, cmov is a loss (as we've already established in this thread) and I think always will be (i.e. cpu vendors seem to be putting more and more smarts into branch prediction instead).
> 

Thanks for the explanations. It’s helpful.


> Yes, that was me not understanding the underlying branch profiling mechanisms.
> 
> Actually, that question of mine was more aimed at John who said we should do something special for loops with max/min accumulators :).
> 

Oh, ok :-)

Paul.
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