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my c++ d c# benchmark!

Kambiz · 2006-10-13T08:18:40

Today I saw the D's pi computing sample (\dmd\samples\d\pi.d) and thought that porting it to c++ and c# for making a little benchmark should be easy: d: import std.c.stdio; import std.c.stdlib; import std.c.time; const int LONG_TIME=4000; byte[] p; byte[] t; int q; int main(char[][] args) { int startime, endtime; int i; if (args.length == 2) { sscanf(&args[1][0],"%d",&q); } else { printf("Usage: pi [precision]\n"); exit(55); } if (q < 0) { printf("Precision was too low, running with precision of 0.\n"); q = 0; } if (q > LONG_TIME) { printf("Be prepared to wait a while...\n"); } // Compute one more digit than we display to compensate for rounding q++; p.length = q + 1; t.length = q + 1; /* compute pi */ std.c.time.time(&startime); arctan(2); arctan(3); mul4(); std.c.time.time(&endtime); // Return to the number of digits we want to display q--; /* print pi */ printf("pi = %d.",cast(int)(p[0])); for (i = 1; i <= q; i++) printf("%d",cast(int)(p)); printf("\n"); printf("%ld seconds to compute pi with a precision of %d digits.\n",endtime-startime,q); return 0; } void arctan(int s) { int n; t[0] = 1; div(s); /* t[] = 1/s */ add(); n = 1; do { mul(n); div(s * s); div(n += 2); if (((n-1) / 2) % 2 == 0) add(); else sub(); } while (!tiszero()); } void add() { int j; for (j = q; j >= 0; j--) { if (t[j] + p[j] > 9) { p[j] += t[j] - 10; p[j-1] += 1; } else p[j] += t[j]; } } void sub() { int j; for (j = q; j >= 0; j--) if (p[j] < t[j]) { p[j] -= t[j] - 10; p[j-1] -= 1; } else p[j] -= t[j]; } void mul(int multiplier) { int b; int i; int carry = 0, digit = 0; for (i = q; i >= 0; i--) { b = (t * multiplier + carry); digit = b % 10; carry = b / 10; t = digit; } } /* t[] /= l */ void div(int divisor) { int i, b; int quotient, remainder = 0; for (i = 0; i <= q; i++) { b = (10 * remainder + t); quotient = b / divisor; remainder = b % divisor; t = quotient; } } void div4() { int i, c, d = 0; for (i = 0; i <= q; i++) { c = (10 * d + p) / 4; d = (10 * d + p) % 4; p = c; } } void mul4() { int i, c, d; d = c = 0; for (i = q; i >= 0; i--) { d = (p * 4 + c) % 10; c = (p * 4 + c) / 10; p = d; } } int tiszero() { int k; for (k = 0; k <= q; k++) if (t[k] != 0) return false; return true; } (I compiled pi.d with dmd -O -release pi.d) c++: #include <stdio.h> #include <stdlib.h> #include <time.h> #define LONG_TIME 4000 __int8* p; __int8* t; int q; void arctan(int s); void add(); void sub(); void mul(int); void div(int); void mul4(); void div4(); int tiszero(); int main(int argv,char** args) { time_t startime, endtime; int i; if (argv == 2) { sscanf(&args[1][0],"%d",&q); } else { printf("Usage: pi [precision]\n"); exit(55); } if (q < 0) { printf("Precision was too low, running with precision of 0.\n"); q = 0; } if (q > LONG_TIME) { printf("Be prepared to wait a while...\n"); } // Compute one more digit than we display to compensate for rounding q++; p = new __int8[ q + 1 ]; t = new __int8[ q + 1 ]; for(int tt=0;tt<=q;tt++) { p=0; t=0; } /* compute pi */ time(&startime); arctan(2); arctan(3); mul4(); time(&endtime); // Return to the number of digits we want to display q--; /* print pi */ printf("pi = %d.",(int)(p[0])); for (i = 1; i <= q; i++) printf("%d",(int)(p)); printf("\n"); printf("%ld seconds to compute pi with a precision of %d digits.\n",(long)endtime-(long)startime,q); delete [] p; delete [] t; return 0; } void arctan(int s) { int n; t[0] = 1; div(s); /* t[] = 1/s */ add(); n = 1; do { mul(n); div(s * s); div(n += 2); if (((n-1) / 2) % 2 == 0) add(); else sub(); } while (!tiszero()); } void add() { int j; for (j = q; j >= 0; j--) { if (t[j] + p[j] > 9) { p[j] += t[j] - 10; p[j-1] += 1; } else p[j] += t[j]; } } void sub() { int j; for (j = q; j >= 0; j--) if (p[j] < t[j]) { p[j] -= t[j] - 10; p[j-1] -= 1; } else p[j] -= t[j]; } void mul(int multiplier) { int b; int i; int carry = 0, digit = 0; for (i = q; i >= 0; i--) { b = (t * multiplier + carry); digit = b % 10; carry = b / 10; t = digit; } } /* t[] /= l */ void div(int divisor) { int i, b; int quotient, remainder = 0; for (i = 0; i <= q; i++) { b = (10 * remainder + t); quotient = b / divisor; remainder = b % divisor; t = quotient; } } void div4() { int i, c, d = 0; for (i = 0; i <= q; i++) { c = (10 * d + p) / 4; d = (10 * d + p) % 4; p = c; } } void mul4() { int i, c, d; d = c = 0; for (i = q; i >= 0; i--) { d = (p * 4 + c) % 10; c = (p * 4 + c) / 10; p = d; } } int tiszero() { int k; for (k = 0; k <= q; k++) if (t[k] != 0) return false; return true; } c#: using System; using System.Diagnostics; class Pi { static void Main(string[] args) { Pi pi = new Pi(); pi.run(args); } private const int LONG_TIME = 4000; sbyte[] p; sbyte[] t; int q; void run(string[] args) { Stopwatch timer = new Stopwatch(); int i; if (args.Length == 1) { q = int.Parse(args[0]); } else { Console.WriteLine("Usage: pi [precision]"); return; } if (q < 0) { Console.WriteLine("Precision was too low, running with precision of 0."); q = 0; } if (q > LONG_TIME) { Console.WriteLine("Be prepared to wait a while..."); } // Compute one more digit than we display to compensate for rounding q++; p = new sbyte[q + 1]; t = new sbyte[q + 1]; /* compute pi */ timer.Start(); arctan(2); arctan(3); mul4(); timer.Stop(); // Return to the number of digits we want to display q--; /* print pi */ Console.Write("pi = {0}.", p[0]); for (i = 1; i <= q; i++) Console.Write(p); Console.WriteLine(); Console.WriteLine("{0} seconds to compute pi with a precision of {1} digits.", timer.ElapsedMilliseconds / 1000.0, q); return; } void arctan(int s) { int n; t[0] = 1; div(s); /* t[] = 1/s */ add(); n = 1; do { mul(n); div(s * s); div(n += 2); if (((n - 1) / 2) % 2 == 0) add(); else sub(); } while (!tiszero()); } void add() { int j; for (j = q; j >= 0; j--) { if (t[j] + p[j] > 9) { p[j] += (sbyte)(t[j] - 10); p[j - 1] += 1; } else p[j] += t[j]; } } void sub() { int j; for (j = q; j >= 0; j--) if (p[j] < t[j]) { p[j] -= (sbyte)(t[j] - 10); p[j - 1] -= 1; } else p[j] -= t[j]; } void mul(int multiplier) { int b; int i; int carry = 0, digit = 0; for (i = q; i >= 0; i--) { b = (t * multiplier + carry); digit = b % 10; carry = b / 10; t = (sbyte)digit; } } /* t[] /= l */ void div(int divisor) { int i, b; int quotient, remainder = 0; for (i = 0; i <= q; i++) { b = (10 * remainder + t); quotient = b / divisor; remainder = b % divisor; t = (sbyte)quotient; } } void div4() { int i, c, d = 0; for (i = 0; i <= q; i++) { c = (10 * d + p) / 4; d = (10 * d + p) % 4; p = (sbyte)c; } } void mul4() { int i, c, d; d = c = 0; for (i = q; i >= 0; i--) { d = (p * 4 + c) % 10; c = (p * 4 + c) / 10; p = (sbyte)d; } } bool tiszero() { int k; for (k = 0; k <= q; k++) if (t[k] != 0) return false; return true; } } I used the release builds for the benchmark, here are the results: d : 26 seconds to compute pi with a precision of 10000 digits. c# : 34,745(34.745) seconds to compute pi with a precision of 10000 digits. c++ : 15 seconds to compute pi with a precision of 10000 digits. I do not want to start a language war... we had enough such wars in the past days and I have just made this benchmark because I saw the “What do you think of the D language?“ thread and was testing d. I'm just surprised a little: I thought that D shouldn't be much slower than c++ and I thought that c# would be much faster. Maybe there is some optimization option I have not used(?) What do you think about the results? (maybe some one can test java) (I really like the invariant and unittest features of D, are there equivalents for c#?) -I can't reply until tomorrow- [Edited by - Kambiz on October 7, 2006 2:02:45 AM]

General and Gameplay Programming Programming Unity

Started by Kambiz October 06, 2006 06:40 PM

70 comments, last by Raghar 17 years, 6 months ago

Raghar

October 13, 2006 08:10 AM

Quote:Original post by Promit
Sure (with the exception of support for vectorized instruction sets, which Java and C# lack).

You could add them yourself. Source code of Java JIT is accessible, and if you don't screw it too much, you could add that little support for SIMD instructions.

BTW SIMD means Single Instruction Multiple Data. And I somehow doubt they would add support for multiple instructions.

Raghar

October 13, 2006 08:18 AM

Quote:Original post by Promit
Sure (with the exception of support for vectorized instruction sets, which Java and C# lack).

You could add them yourself. Source code of Java JIT is accessible, and if you don't screw it too much, you could add that little support for SIMD instructions.
The problem you'd know if you'd use ASM is SIMD registers are not that great win. First at all, they are small 128 bits is VERY small if you need to work with 64 bit numbers.
Suport for possibility to make all operations on 128 bit regiseters in one clock cycle was added by introducing Core2Duo. Previous CPUs splited some operations/all? into two 64 bit operations.

BTW SIMD means Single Instruction Multiple Data. And I somehow doubt they would add support for Multiple Instructions...

my c++ d c# benchmark!

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my c++ d c# benchmark!

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