一、基本信息
實(shí)踐題目：同步對(duì)象實(shí)現(xiàn)P、V操作
二、實(shí)踐內(nèi)容簡(jiǎn)要描述
實(shí)踐目標(biāo)：
用Win32提供的同步對(duì)象實(shí)現(xiàn)P、V操作，并解決任務(wù)4中有限緩沖區(qū)問(wèn)題
實(shí)踐內(nèi)容：
用同步對(duì)象實(shí)現(xiàn)P、V操作解決一下問(wèn)題：
寫一個(gè)線程實(shí)現(xiàn)C/C++語(yǔ)言程序：一些線程負(fù)責(zé)找出某個(gè)數(shù)據(jù)范圍的素?cái)?shù)，并放到一個(gè)數(shù)組中，另一些線程負(fù)責(zé)將數(shù)組中的素?cái)?shù)按次序取出，并顯示出來(lái)。顯示過(guò)的數(shù)據(jù)要從數(shù)組中刪除。要求定義一個(gè)全局變量的數(shù)組：int prime[9]用于存放找到的待顯示的素?cái)?shù)。
三、實(shí)踐報(bào)告主要內(nèi)容
設(shè)計(jì)思路：
實(shí)現(xiàn)P、V操作用了binary semaphore來(lái)實(shí)現(xiàn)，
wait操作的實(shí)現(xiàn)
 Wait(s1);
 c--;
if(c<0){
 signal(s1);
 wait(s2);
}
signal(s1);
 signal操作的實(shí)現(xiàn)：
wait(s1);
c++;
if(c<=0) 
 signal(s2);
else
 signal(s1);
 線程同步仍采用了生產(chǎn)者－消費(fèi)者算法。
P、V操作用了一個(gè)返回值，為0時(shí)表示超時(shí)，為1表示等待成功。
主要數(shù)據(jù)結(jié)構(gòu)：
線程信息結(jié)構(gòu)體：
int searchCount = 0;  //已開線程數(shù)
int displayCount = 0;  //已開線程數(shù)

CRITICAL_SECTION  my_section;

HANDLE h_semaphore_s1_e = CreateSemaphore(NULL, 1, 1, "Bin_S1_Empty"); //Empty的S1
HANDLE h_semaphore_s2_e = CreateSemaphore(NULL, 0, 1, "Bin_S2_Empty"); //Empty的S2

HANDLE h_semaphore_s1_f = CreateSemaphore(NULL, 1, 1, "Bin_S1_Full"); //Full的S1
HANDLE h_semaphore_s2_f = CreateSemaphore(NULL, 0, 1, "Bin_S2_Full"); //Full的S2

int Count_Empty;     //Empty計(jì)數(shù)器
int Count_Full;     //Full計(jì)數(shù)器
struct ThreadInfo
{
 int   serial;
 char  command;
 int   num_from;
 int   num_to;
};
int prime[9];
void GetCommand( char* file );
void thd_display(void* p);
void thd_search(void* p);   
BOOL IsPrime(int num);   //用于判斷是否為素?cái)?shù)
int  GetUsed(int num[9]);   //用于獲得剩余空間

void P(LPCTSTR lpcstr,int &count);  //P 操作
void V(LPCTSTR lpcstr,int &count);  //V 操作
主要代碼分析：
int wait(int *,HANDLE Event[]);  // binary semaphore實(shí)現(xiàn)的P 操作
int signal(int *,HANDLE Event[]); // binary semaphore實(shí)現(xiàn)的V 操作
四、實(shí)踐結(jié)果
基本數(shù)據(jù)：
 
源程序代碼行數(shù) 完成實(shí)踐投入的總時(shí)間 資料查閱時(shí)間 編程調(diào)試時(shí)間 源程序文件    
413 1小時(shí) 0.5小時(shí) 0.5小時(shí) ex5.cpp  
測(cè)試數(shù)據(jù)設(shè)計(jì)：
1.“ex4.dat”文件內(nèi)容測(cè)試結(jié)果同ex4：
2.自設(shè)測(cè)試數(shù)據(jù)
 1  w  10  100
2  w  1   10
3  W  2   20
4  D  20   0
5  D   3   0
測(cè)試結(jié)果分析：
對(duì)應(yīng)于數(shù)據(jù)的運(yùn)行結(jié)果：
Reader Priority:

Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Search thread 2 begins to write the date .
Search thread 2 finish writing the date .
Search thread 3 begins to write the date .
Search thread 3 finish writing the date .
Search thread 2 begins to write the date .
Search thread 2 finish writing the date .
Search thread 3 begins to write the date .
Search thread 3 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "11".
Display thread 4 finishes outputting prime num .
Display thread 5 begins to output prime num .
----------Display thread 5 output the prime num "13".
Display thread 5 finishes outputting prime num .
Search thread 2 begins to write the date .
Search thread 2 finish writing the date .
Search thread 3 begins to write the date .
Search thread 3 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "7".
Display thread 4 finishes outputting prime num .
Display thread 5 begins to output prime num .
----------Display thread 5 output the prime num "7".
Display thread 5 finishes outputting prime num .
Search thread 3 begins to write the date .
Search thread 3 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "11".
Display thread 4 finishes outputting prime num .
Display thread 5 begins to output prime num .
----------Display thread 5 output the prime num "17".
Display thread 5 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Search thread 3 begins to write the date .
Search thread 3 finish writing the date .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Search thread 3 begins to write the date .
Search thread 3 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "23".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "31".
Display thread 4 finishes outputting prime num .
Search thread 3 begins to write the date .
Search thread 3 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "19".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "37".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "41".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "43".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "47".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "53".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "59".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "61".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "67".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "71".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "73".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "79".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "83".
Display thread 4 finishes outputting prime num .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "13".
Display thread 4 finishes outputting prime num .
Display thread 4 begins to output prime num .
----------Display thread 4 output the prime num "29".
Display thread 4 finishes outputting prime num .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
Search thread 1 begins to write the date .
Search thread 1 finish writing the date .
All search and display thread have finished Operating.
The array at last.
 89  97  0  19  3  3  5  5  17

Press any key to finish this Program.
Thank you test this Program!
程序運(yùn)行過(guò)程中，線程1、線程2、線程3為找素?cái)?shù)線程，線程4、線程5負(fù)責(zé)顯示，它們均交叉訪問(wèn)臨界區(qū)，程序運(yùn)行結(jié)果完全正確。
五、實(shí)踐體會(huì)
 該實(shí)驗(yàn)與實(shí)驗(yàn)四相比，只是用軟件的方法來(lái)實(shí)現(xiàn)內(nèi)核對(duì)象的功能，只須細(xì)心就能順利完成任務(wù)。
 通過(guò)實(shí)驗(yàn)我基本了解了P、V操作的具體實(shí)現(xiàn)，以及有限緩沖區(qū)問(wèn)題的解決方法。
七、參考文獻(xiàn)
1、MSDN
2、《操作系統(tǒng)概念（第六版 影印版）》，Abraham Silberschatz、Peter Baer Galvin、Greg Gagne，
高等教育出版社

附：源程序部分關(guān)鍵代碼
void P(LPCTSTR lpcstr,int &count)
{
 HANDLE h_s1,h_s2;
 if(strcmp(lpcstr,"Empty") == 0)
 {
  h_s1=OpenSemaphore(SEMAPHORE_ALL_ACCESS,FALSE,"Bin_S1_Empty");
  h_s2=OpenSemaphore(SEMAPHORE_ALL_ACCESS,FALSE,"Bin_S2_Empty");
 }
 else if(strcmp(lpcstr,"Full") == 0)
 {
  h_s1=OpenSemaphore(SEMAPHORE_ALL_ACCESS,FALSE,"Bin_S1_Full");
  h_s2=OpenSemaphore(SEMAPHORE_ALL_ACCESS,FALSE,"Bin_S2_Full");
 }
 
 WaitForSingleObject(h_s1, INFINITE);

 count--;

 if (count < 0)
 {
  ReleaseSemaphore(h_s1, 1, NULL);
  WaitForSingleObject(h_s2, INFINITE);
 }

 ReleaseSemaphore(h_s1, 1, NULL);

 CloseHandle(h_s1);
 CloseHandle(h_s2);

}

void V(LPCTSTR lpcstr,int &count)
{
 HANDLE h_s1,h_s2;
 if(strcmp(lpcstr,"Empty") == 0)
 {
  h_s1=OpenSemaphore(SEMAPHORE_ALL_ACCESS,FALSE,"Bin_S1_Empty");
  h_s2=OpenSemaphore(SEMAPHORE_ALL_ACCESS,FALSE,"Bin_S2_Empty");
 }
 else if(strcmp(lpcstr,"Full") == 0)
 {
  h_s1=OpenSemaphore(SEMAPHORE_ALL_ACCESS,FALSE,"Bin_S1_Full");
  h_s2=OpenSemaphore(SEMAPHORE_ALL_ACCESS,FALSE,"Bin_S2_Full");
 }
 
 WaitForSingleObject(h_s1, INFINITE);

 count++;

 if (count <= 0)
 {
  ReleaseSemaphore(h_s2, 1, NULL);
 }
 else
 {
  ReleaseSemaphore(h_s1, 1, NULL);
 }
 CloseHandle(h_s1);
 CloseHandle(h_s2);

}

任務(wù)六  LINUX下以原生方式查看軟盤信息
一、基本信息
實(shí)踐題目：在LINUX下以原生（raw）方式查看軟盤的信息
二、實(shí)踐內(nèi)容簡(jiǎn)要描述
實(shí)踐目標(biāo)：
熟悉Linux操作系統(tǒng)下的基本操作及其程序設(shè)計(jì)的基本方法。
學(xué)會(huì)在LINUX下使用原生方式查看軟盤上的數(shù)據(jù)信息。
實(shí)踐內(nèi)容：
在LINUX下使用原生方式查詢并顯示目前磁盤的以下信息：
－每扇區(qū)字節(jié)數(shù)
－每磁道扇區(qū)數(shù)
－每柱面的磁道數(shù)（磁頭數(shù)）
三、實(shí)踐報(bào)告主要內(nèi)容
設(shè)計(jì)思路：
只要正確定位文件的讀取位置并確定所需讀取的字節(jié)數(shù)便可以正確讀取軟盤上的數(shù)據(jù)，先用open函數(shù)以只讀方式打開軟盤，再用lseek定位讀取位置，然后用read讀取相應(yīng)字節(jié)的數(shù)據(jù)即可。
主要數(shù)據(jù)結(jié)構(gòu)：
 int fd;     //文件描述符
 size_t num_read;
 int bytes_per_sector = 0;  //每扇區(qū)字節(jié)數(shù)
 int sectors_per_track = 0;  //每磁道扇區(qū)數(shù)
 int number_of_surface = 0;  //每柱面的磁道數(shù)（磁頭數(shù)）
 int * pi;  //指向文件緩沖區(qū)的指針
 pi = &bytes_per_sector;
 pi = &sectors_per_track;
 pi = &number_of_surface;
主要代碼分析： 詳見(jiàn)代碼注釋。
四、實(shí)踐結(jié)果
基本數(shù)據(jù)：
 
源程序代碼行數(shù) 完成實(shí)踐投入的總時(shí)間 資料查閱時(shí)間 編程調(diào)試時(shí)間 源程序文件    
43 0.5小時(shí) 0.4小時(shí) 0.1小時(shí) ex6.cpp  
測(cè)試數(shù)據(jù)設(shè)計(jì)：
3.5寸軟盤一張。
測(cè)試結(jié)果分析：
程序運(yùn)行結(jié)果為：
Bytes per sector: 512
Sectors per track: 18
Number of surfaces:: 2
程序運(yùn)行結(jié)果完全正確。
五、實(shí)踐體會(huì)
 以原生方式讀取磁盤信息時(shí)，將磁盤當(dāng)作一個(gè)文件打開，要正確定位文件的讀取位置，并確定讀取字節(jié)數(shù)，這樣便能從中讀取相應(yīng)數(shù)據(jù)。通過(guò)實(shí)驗(yàn)，了解了軟盤上BOOT的布局，并熟悉了Linux操作系統(tǒng)下程序設(shè)計(jì)的基本方法和基本操作。
六、參考文獻(xiàn)
1、《Linux基礎(chǔ)教程》，黃慶生 吳振華 蘇哲 編著，人民郵電出版社
附：源程序代碼
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <unistd.h>
#include <stdio.h>

int main()
{
 //每扇區(qū)字節(jié)數(shù)   0x0b-0x0c  11 - 12
 //每磁道扇區(qū)數(shù)   0x18-0x19  24 - 25
 //每柱面的磁道數(shù) 0x1a-0x1b  26 – 27

 int fd;
 size_t num_read;
 int bytes_per_sector   = 0;
 int sectors_per_track  = 0;
 int number_of_surface  = 0;

 int * pi;
 fd = open("/dev/fd0", O_RDONLY);

 pi = &bytes_per_sector;
 lseek(fd,(off_t)11,SEEK_SET);
 num_read = read(fd,(void *)pi,(size_t)2);
 printf("Bytes per sector: %d\n",bytes_per_sector);

 pi = &sectors_per_track;
 lseek(fd,(off_t)24,SEEK_SET);
 num_read = read(fd,(void *)pi,(size_t)2);
 printf("Sectors per track: %d\n",sectors_per_track);

 pi = &number_of_surface;
 lseek(fd,(off_t)26,SEEK_SET);
 num_read = read(fd,(void *)pi,(size_t)2);
 printf("Number of surfaces: %d\n",number_of_surface);
 return 0; 
}