Comparision Among The Access Speed Of Cache , Memory And Disk

时间：2010-09-17 来源：坑王

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 9000
12  #define CACHE_BLOCK_SIZE 64
13  #define MEMORY_TIMES 1000
14  #define DISK_TIMES 100
15 #define MEMORY_BLOCK_SIZE 4096
16 
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