Date: Thu, 7 Mar 2002 10:45:18 +0200 From: Dimitar Peikov <mitko@rila.bg> To: freebsd-hackers@freebsd.org Subject: Swapping performance Message-ID: <20020307104518.0f73740b.mitko@rila.bg>
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[-- Attachment #1 --] I start some performance tests on -stable and on SuSE 7.1 / 2.4.17. I don't comment about 'bzero' performance, but when RAM is over, Linux is much faster. I have no idea what is the algorithm of swapping but it seems that the granularity of swapping pieces is the key or the importance of swapping memory blocks of certain task. Ooo I forgot to say that the both machines have the same hardware, IBM 300PL, 256 RAM and no other tasks running. I had to run these tests to choose the fastest platform for building our software indexes, which requires a lot of math and memory operations. --- with bzero --- Linux$ time ./malloc_test *# real 0m37.640s user 0m1.370s sys 0m2.950s Linux$ FreeBSD$ time ./malloc_test *# real 0m46.640s user 0m2.280s sys 0m2.550s FreeBSD$ --- without bzero --- Linux$ time ./malloc_test *# real 0m6.371s user 0m0.450s sys 0m1.510s Linux$ FreeBSD$ time ./malloc_test *# real 0m11.571s user 0m1.150s sys 0m1.830s FreeBSD$ -- Dimitar Peikov Programmer Analyst Globalization Group "We Build e-Business" RILA Solutions 27 Building, Acad.G.Bonchev Str. 1113 Sofia, Bulgaria phone: (+359 2) 9797320 phone: (+359 2) 9797300 fax: (+359 2) 9733355 http://www.rila.com [-- Attachment #2 --] #include <stdlib.h> #include <string.h> #include <stdio.h> #define MALLOC_SIZE 1024*1024*256 int main(int argc, char **argv) { char *ptr; int i, i_count; int j; ptr = (char *) malloc(MALLOC_SIZE); bzero(ptr, MALLOC_SIZE); i_count = MALLOC_SIZE / 16; fprintf(stderr, "*"); for (i = 0; i < i_count; i ++) { ptr[i >> 4] = ptr[(i >> 3) + 1]++; } fprintf(stderr, "#"); for (j = 0; j < i_count; j ++) { ptr[j << 4] = ptr[(j >> 3) + 1]++; } free(ptr); return 0; }
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