Comparision Among The Access Speed Of Cache , Memory And Disk

The first homework of db design is to compare the speed of accessing cache , meory and disk. This experiment show the gap among them.

To test the accecc time of cache , I use cpu-z to find that the block size in my cpu cache is 64 byte. That means I can create a 64-byte-size array and access it repeatly to estimate the accecc time.

The block size in my main memory with win7 is 4k. So i make a array and access it every 4k byte sequentially.

To obtain the time of accessing disk, a large file will be readed every at intervals of 4k byte.

here i show the c++ code:

1 // Compare.cpp : 定义控制台应用程序的入口点。 2  // 3   4 #include "stdafx.h" 5 #include <ctime> 6 #include <iostream> 7 #include <fstream> 8  9  using namespace std;10 11  #define CACHE_TIMES 900012  #define CACHE_BLOCK_SIZE 6413  #define MEMORY_TIMES 100014  #define DISK_TIMES 10015 #define MEMORY_BLOCK_SIZE 409616 17 clock_t  start, stop; /* clock_t is a built-in type for processor time (ticks) */18 double  duration;  /* records the run time (seconds) of a function */19 20 void  TestCache ()21 {22     char test[CACHE_TIMES];23     start = clock();24     25     for (int i = 0 ; i < CACHE_TIMES*1000 ; i++)26     {27         test[i%CACHE_BLOCK_SIZE]++;   28     }29 30     stop = clock();31     duration = ((double)(stop - start)*1000000)/(CLK_TCK*CACHE_TIMES); 32     cout << "The time to access cache is  : " << duration <<"ns"<< endl;33 }34 35 void TestMemory()36 {37     char test[MEMORY_TIMES*CACHE_BLOCK_SIZE];    38     char dat;39     start = clock();40     for (int i = 0 ; i < MEMORY_TIMES*1000 ; i++)41     {42         dat = test[(i*CACHE_BLOCK_SIZE)%(MEMORY_TIMES*CACHE_BLOCK_SIZE)];   43     }44     stop = clock();45     duration = ((double)(stop - start)*1000)/(CLK_TCK*MEMORY_TIMES);     46     cout << "The time to access memory is  : " << duration <<"us"<< endl;47 }48 49 void TestDisk()50 {51     ofstream test("test.dat");52     char dat;53     char buf[20];54     for(int i = 0 ; i < DISK_TIMES*MEMORY_BLOCK_SIZE ; i ++)55     {56         dat = i % 256;57         test << dat;58     }59     test.close();60 61     FILE *fp;62 63     start = clock();64     fp=fopen("test.dat","r");65     for (long i = 0 ; i < DISK_TIMES ; i++)66     {67         fseek(fp,i*MEMORY_BLOCK_SIZE,SEEK_SET);    68         fread (buf , 1 , 1, fp);69     }70     fclose(fp);71     stop = clock();72     duration = ((double)(stop - start)*1000)/(CLK_TCK*DISK_TIMES);     73     cout << "The time to access disk is  : " <<duration <<"ms"<< endl;74 75 }76 77 int _tmain(int argc, _TCHAR* argv[])78 {79     TestCache();80     TestMemory();81     TestDisk();82     getchar();83     return 0;84 }85 86

result:

The time to access cache is : 4.88889ns

The time to access memory is : 0.014us

The time to access disk is : 0.01ms
原文链接: https://www.cnblogs.com/kking/archive/2010/09/17/1829607.html

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