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Dec 6, 2016 at 7:36 comment added slang This will tell you far more about the version of bash or type of shell being used than the performance of your CPU.
Jan 29, 2016 at 16:39 comment added Steven Lu This is not 6 gigaflops, this is ~0.17 megabashincrements. And, my macbook owns all: ( i=0 ; while (( i < 1000000 )); do; (( i ++ )); done; ) 1.33s user 0.00s system 99% cpu 1.337 total
Mar 9, 2015 at 12:58 comment added JohnMudd Good point, I should have updated this earlier. Here's my current approach: time ( i=0; while (( i < 1000000 )); do (( i ++ )); done )
Mar 7, 2015 at 21:23 comment added xenithorb Quicker way to get to the number you're really after, i.e. seconds that have passed: DATE=$(date +%s); i=0; while (( i < 1000000 )); do (( i ++ )); done; echo $(( $(date +%s)-DATE ))
Jul 16, 2014 at 15:29 comment added JohnMudd The general goal was "measure cpu performance". Later mentions Gflops but I think that's actually a misguided tangent. If Gflops was the true goal then asking about CPU performance was a mistake.
Jul 15, 2014 at 16:58 comment added Anthon Would that be 43 - 37 = 6 GigaFlops? I fail to see how an integer increment in a shell, relates to floating point performance.
Jul 15, 2014 at 15:49 review Low quality posts
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Jul 15, 2014 at 15:34 review First posts
Jul 15, 2014 at 15:40
Jul 15, 2014 at 15:30 history answered JohnMudd CC BY-SA 3.0