银行家算法c++软编码
时间: 2023-11-26 08:49:03 浏览: 30
根据提供的引用内容,我们可以了解到银行家算法是一种资源分配算法,用于避免死锁和保证系统安全。下面是一个简单的C++实现:
```c++
#include <iostream>
using namespace std;
const int P = 5; // 进程数
const int R = 3; // 资源数
int available[R] = {3, 3, 2}; // 可用资源数
int maxm[P][R] = {{7, 5, 3}, {3, 2, 2}, {9, 0, 2}, {2, 2, 2}, {4, 3, 3}}; // 最大需求矩阵
int allocation[P][R] = {{0, 1, 0}, {2, 0, 0}, {3, 0, 2}, {2, 1, 1}, {0, 0, 2}}; // 已分配矩阵
int need[P][R]; // 需求矩阵
bool isSafe() {
int work[R];
bool finish[P] = {false};
// 初始化work数组
for (int i = 0; i < R; i++) {
work[i] = available[i];
}
// 找到一个未完成的进程,且它所需的资源小于等于work数组中可用的资源
for (int i = 0; i < P; i++) {
if (finish[i] == false) {
bool flag = true;
for (int j = 0; j < R; j++) {
if (need[i][j] > work[j]) {
flag = false;
break;
}
}
if (flag) {
// 将该进程所占用的资源释放,并将其标记为已完成
for (int j = 0; j < R; j++) {
work[j] += allocation[i][j];
}
finish[i] = true;
i = -1; // 重新开始循环
}
}
}
// 如果所有进程都已完成,则说明系统是安全的
for (int i = 0; i < P; i++) {
if (finish[i] == false) {
return false;
}
}
return true;
}
int main() {
// 计算need矩阵
for (int i = 0; i < P; i++) {
for (int j = 0; j < R; j++) {
need[i][j] = maxm[i][j] - allocation[i][j];
}
}
// 模拟进程申请资源
int request[P][R] = {{0, 0, 0}, {2, 0, 2}, {0, 0, 0}, {1, 0, 0}, {0, 0, 2}};
for (int i = 0; i < P; i++) {
// 如果该进程所需的资源小于等于可用资源,则分配资源
bool flag = true;
for (int j = 0; j < R; j++) {
if (request[i][j] > available[j]) {
flag = false;
break;
}
}
if (flag) {
for (int j = 0; j < R; j++) {
available[j] -= request[i][j];
allocation[i][j] += request[i][j];
need[i][j] -= request[i][j];
}
if (isSafe()) {
cout << "进程" << i << "申请资源成功,分配后的资源情况为:" << endl;
cout << "可用资源数:";
for (int j = 0; j < R; j++) {
cout << available[j] << " ";
}
cout << endl;
cout << "已分配矩阵:" << endl;
for (int j = 0; j < P; j++) {
for (int k = 0; k < R; k++) {
cout << allocation[j][k] << " ";
}
cout << endl;
}
cout << endl;
} else {
// 如果分配后不安全,则回滚操作
for (int j = 0; j < R; j++) {
available[j] += request[i][j];
allocation[i][j] -= request[i][j];
need[i][j] += request[i][j];
}
cout << "进程" << i << "申请资源失败,分配后不安全" << endl;
}
} else {
cout << "进程" << i << "申请资源失败,所需资源大于可用资源" << endl;
}
}
return 0;
}
```