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``` name old time/op new time/op delta Buzhash2/1K-4 610ns ± 4% 643ns ±16% ~ (p=0.421 n=8+10) Buzhash2/1M-4 1.25ms ± 5% 1.16ms ± 4% -7.31% (p=0.000 n=10+10) Buzhash2/16M-4 19.2ms ± 2% 17.5ms ± 2% -8.73% (p=0.000 n=9+9) Buzhash2/100M-4 117ms ± 1% 107ms ± 3% -8.26% (p=0.000 n=10+10) name old speed new speed delta Buzhash2/1K-4 1.68GB/s ± 4% 1.60GB/s ±14% ~ (p=0.408 n=8+10) Buzhash2/1M-4 842MB/s ± 5% 908MB/s ± 3% +7.86% (p=0.000 n=10+10) Buzhash2/16M-4 875MB/s ± 2% 959MB/s ± 2% +9.57% (p=0.000 n=9+9) Buzhash2/100M-4 897MB/s ± 1% 977MB/s ± 3% +9.02% (p=0.000 n=10+10) name old alloc/op new alloc/op delta Buzhash2/1K-4 1.17kB ± 1% 1.17kB ± 0% -0.50% (p=0.006 n=10+10) Buzhash2/1M-4 1.08MB ± 1% 1.07MB ± 0% ~ (p=0.739 n=10+10) Buzhash2/16M-4 17.1MB ± 0% 17.1MB ± 0% ~ (p=0.579 n=10+10) Buzhash2/100M-4 106MB ± 0% 106MB ± 0% -0.01% (p=0.000 n=9+7) name old allocs/op new allocs/op delta Buzhash2/1K-4 3.00 ± 0% 3.00 ± 0% ~ (all equal) Buzhash2/1M-4 8.00 ± 0% 8.00 ± 0% ~ (all equal) Buzhash2/16M-4 72.0 ± 0% 72.0 ± 0% ~ (all equal) Buzhash2/100M-4 406 ± 0% 406 ± 0% ~ (all equal) ``` License: MIT Signed-off-by: Jakub Sztandera <kubuxu@protocol.ai>
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@Kubuxu I am not familiar with this idiom and googling does not turn up anything. What does looking at a value at an index and throwing it away achieve?