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Why don't all float functions have t= he=0A= > same workload?=0A= =0A= Switching off the workload feature doesn't make it possible to compare perf= ormance=0A= between different math functions - the input files are different anyway.=0A= =0A= What we want to know is how each math function performs on typical inputs.= =0A= The workloads contain traces from real applications and the workload featur= e=0A= processes these traces in a realistic way to get accurate throughput and la= tency=0A= results. So the datasets are unique for each math function.=0A= =0A= The performance results of the workloads can be compared since it is a meas= ure=0A= how each function performs on typical inputs. Comparing performance of sinf= (0)=0A= with logf(0) on the other hand is useless.=0A= =0A= Cheers,=0A= Wilco=