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| Subject | Basic benchmark |
| From | bearophile <bearophileHUGS@lycos.com> |
| Date | Sat, 13 Dec 2008 10:21:53 -0500 |
| Newsgroups | digitalmars.D |
I have adapted another small benchmark to D. This benchmark is less interesting than the other ones because it mostly tests the optimizations done by the back-end. This means it's not a problem of the D language or its front-end, so even if DMD here shows to be not much efficient, LDC once finished may show significant improvements. As usual I may have done several errors, so keep your eyes open. D code: /* original code copyright 2004 Christopher W. Cowell-Shah http://www.cowell-shah.com/research/benchmark/code other code portions copyright http://dada.perl.it/shootout/ and Doug Bagley http://www.bagley.org/~doug/shootout combined, modified and fixed by Thomas Bruckschlegel - http://www.tommti-systems.com */ import std.c.stdio: printf; import std.c.time: clock, CLOCKS_PER_SEC, clock_t; void longArithmetic(long longMin, long longMax) { clock_t startTime = clock(); long longResult = longMin; long i = longMin; while (i < longMax) { longResult -= i++; longResult += i++; longResult *= i++; longResult /= i++; } clock_t stopTime = clock(); double elapsedTime = (stopTime - startTime) / (CLOCKS_PER_SEC / 1000.0); printf("Long arithmetic elapsed time: %1.0f ms with longMax %ld\n", elapsedTime, longMax); printf(" i: %ld\n", i); printf(" longResult: %ld\n", longResult); } void nested_loops(int n) { clock_t startTime = clock(); int a, b, c, d, e, f; int x=0; for (a=0; a<n; a++) for (b=0; b<n; b++) for (c=0; c<n; c++) for (d=0; d<n; d++) for (e=0; e<n; e++) for (f=0; f<n; f++) x+=a+b+c+d+e+f; clock_t stopTime = clock(); double elapsedTime = (stopTime - startTime) / (CLOCKS_PER_SEC / 1000.0); printf("Nested Loop elapsed time: %1.0f ms %d\n", elapsedTime, x); } int main() { long longMin = 10_000_000_000L; long longMax = 11_000_000_000L; longArithmetic(longMin, longMax); nested_loops(40); return 0; } ------------------------ C code, almost the same (you may need to change LL_FORMAT to make it run correctly): /* original code copyright 2004 Christopher W. Cowell-Shah http://www.cowell-shah.com/research/benchmark/code other code portions copyright http://dada.perl.it/shootout/ and Doug Bagley http://www.bagley.org/~doug/shootout combined, modified and fixed by Thomas Bruckschlegel - http://www.tommti-systems.com */ #include "time.h" #include "stdio.h" // accopding to your compiler #define LL_FORMAT "%I64d" //#define LL_FORMAT "%ld" void longArithmetic(long long longMin, long long longMax) { clock_t startTime = clock(); long long longResult = longMin; long long i = longMin; while (i < longMax) { longResult -= i++; longResult += i++; longResult *= i++; longResult /= i++; } clock_t stopTime = clock(); double elapsedTime = (stopTime - startTime) / (CLOCKS_PER_SEC / 1000.0); printf("Long arithmetic elapsed time: %1.0f ms with longMax "LL_FORMAT"\n", elapsedTime, longMax); printf(" i: "LL_FORMAT"\n", i); printf(" longResult: "LL_FORMAT"\n", longResult); } void nested_loops(int n) { clock_t startTime = clock(); int a, b, c, d, e, f; int x=0; for (a=0; a<n; a++) for (b=0; b<n; b++) for (c=0; c<n; c++) for (d=0; d<n; d++) for (e=0; e<n; e++) for (f=0; f<n; f++) x+=a+b+c+d+e+f; clock_t stopTime = clock(); double elapsedTime = (stopTime - startTime) / (CLOCKS_PER_SEC / 1000.0); printf("Nested Loop elapsed time: %1.0f ms %d\n", elapsedTime, x); } int main() { long long longMin = 10000000000LL; long long longMax = 11000000000LL; longArithmetic(longMin, longMax); nested_loops(40); return 0; } ------------------- I have compiled it with GCC and DMD with: gcc version 4.2.1-dw2 (mingw32-2) -O3 -s DMD v1.037 -O -release -inline --------------------- Timings: C gcc: Long arithmetic: 11.15 s Nested Loops: 0.11 s D dmd: Long arithmetic: 63.7 s Nested Loops: 6.17 s Bye, bearophile | |
Basic benchmark (Current message)
