东南大学操作系统实验报告--银行家算法(共17页)

1、精选优质文档-倾情为你奉上操作系统实验三：银行家算法的实现一、 基本信息：a) 实验题目:银行家算法的实现b) 完成人姓名：韩璐璐c) 学号：d) 报告日期：2016.5.27二、 实验目的通过实验，加深对多实例资源分配系统中死锁避免方法银行家算法的理解，掌握Windows环境下银行家算法的实现方法，同时巩固利用Windows API进行共享数据互斥访问和多线程编程的方法。三、 实验内容1. 在Windows操作系统上，利用Win32 API编写多线程应用程序实现银行家算法。2. 创建n个线程来申请或释放资源，只有保证系统安全，才会批准资源申请。3. 通过Win32 API提供的信号量机制，实

2、现共享数据的并发访问。四、程序运行时的初值和运行结果（系统截图）五、源程序并附上注释#include <iostream>#include <time.h>#include <vector>#include <windows.h>using namespace std;int r3 = 3, 3, 2 ;/系统拥有的资源int r0 = 0, r1 = 0, r2 = 0;/记录申请资源class pcbpublic:int id;bool state;int max3;int alc3;int need3;pcb()void init()sta

3、te = false;cout << "请输入进程的id，各个资源总需求量和已占用资源" << endl;cin >> id;cout << "a,b,c三种资源的最大使用量" << endl;cin >> max0 >> max1 >> max2;cout << "a,b,c三种资源的已占有量" << endl;cin >> alc0 >> alc1 >> alc2;int rd

4、(int n)return rand() % (n + 1);int request()/Sleep(1000);r0 = rd(max0 - alc0);r1 = rd(max1 - alc1);r2 = rd(max2 - alc2);cout << "进程" << id << "申请资源a" << " " << r0 << " 申请资源b" << " " << r1 << "

5、; 申请资源c" << " " << r2 << endl;if (r0>r0 | r1>r1 | r2>r2)cout << "没有那么多资源!" << endl;return 0;if (r0 = (max0 - alc0) && r1 = (max1 - alc1) && r2 = (max2 - alc2)r0 = r0 + alc0;r1 = r1 + alc1;r2 = r2 + alc2;return 1;return 2

6、;bool safe(vector <pcb> temp, int i)int u = r0 - r0, k = r1 - r1, l = r2 - r2;for (int j = i; j<temp.size() - 1; j+)tempj = tempj + 1;temp.pop_back();int size = temp.size();/记录下容器内还有多少个进程/int rangesize;/记录下队列int x = 0;/计数器while (!temp.empty()static int j = 0;if (tempj.max0 - tempj.alc0) <

7、;= u && (tempj.max1 - tempj.alc1) <= k &&(tempj.max2 - tempj.alc2) <= l)/判断是否能运行完cout << "运行" << tempj.id << endl;u = u + tempj.alc0;k = k + tempj.alc1;l = l + tempj.alc2;for (int e = j; e<temp.size() - 1; e+)tempe = tempe + 1;temp.pop_back();if (

8、j >= temp.size()j = 0;elsej+;if (j >= temp.size()j = 0;x+;if (x = (size*size)cout << "没有安全队列,以上情况不成立" << endl;cout << endl;return false;return true;int main()srand(time(0);pcb p4;vector <pcb> vp;for (int i = 0; i<4; i+)pi.init();vp.push_back(pi);int x = 0;/

9、计算器int c;cout << "请选择分配资源方法:1.银行家算法 2.随机算法" << endl;cin >> c;switch (c)case 1:while (!vp.empty()int a;static int i = 0;if (a = vpi.request() != 0)if (a = 1)cout << "进程" << vpi.id << "已经结束" << endl;for (int j = i; j<vp.size()

10、- 1; j+)vpj = vpj + 1;vp.pop_back();if (i >= vp.size()i = 0;cout << "a资源还剩" << r0 << " b资源还剩" << r1 << " c资源还剩" << r2 << endl;cout << endl;elseif (safe(vp, i)cout << "存在安全队列" << endl;cout <<

11、endl;vpi.alc0 = vpi.alc0 + r0;vpi.alc1 = vpi.alc1 + r1;vpi.alc2 = vpi.alc2 + r2;r0 = r0 - r0;r1 = r1 - r1;r2 = r2 - r2;cout << "a资源还剩" << r0 << " b资源还剩" << r1 << " c资源还剩" << r2 << endl;cout << endl;i+;if (i >= vp.size()

12、i = 0;elsei+;if (i >= vp.size()i = 0;x+;if (x >= 200)cout << "初始化的表不安全" << endl;return 0;cout << "进程已经全部结束" << endl;break;case 2:while (!vp.empty()int a2;static int i2 = 0;if (a2 = vpi2.request() != 0)if (a2 = 1)cout << "进程" <<

13、vpi2.id << "已经结束" << endl;for (int j = i2; j<vp.size() - 1; j+)vpj = vpj + 1;vp.pop_back();if (i2 >= vp.size()i2 = 0;cout << "a资源还剩" << r0 << " b资源还剩" << r1 << " c资源还剩" << r2 << endl;cout << end

14、l;elsevpi2.alc0 = vpi2.alc0 + r0;vpi2.alc1 = vpi2.alc1 + r1;vpi2.alc2 = vpi2.alc2 + r2;r0 = r0 - r0;r1 = r1 - r1;r2 = r2 - r2;cout << "a资源还剩" << r0 << " b资源还剩" << r1 << " c资源还剩" << r2 << endl;cout << endl;i2+;if (i2 >= vp.size()i2 = 0;elsei2+;if (i2 >= vp.size()i2 = 0;x+;if (x >= 200)cout << "产生死锁" << endl;return 0;cout << "进程已经全部结束" << endl;b