c语言银行家算法运行实例

以下是一个简单的银行家算法的实现示例，使用 C 语言编写：

#include <stdio.h>
#include <stdbool.h>

#define NUM_RESOURCES 3
#define NUM_PROCESSES 5

int available[NUM_RESOURCES] = {3, 3, 2};
int max_need[NUM_PROCESSES][NUM_RESOURCES] = {
    {7, 5, 3},
    {3, 2, 2},
    {9, 0, 2},
    {2, 2, 2},
    {4, 3, 3},
};
int allocated[NUM_PROCESSES][NUM_RESOURCES] = {
    {0, 1, 0},
    {2, 0, 0},
    {3, 0, 2},
    {2, 1, 1},
    {0, 0, 2},
};

// 计算已分配资源和需要资源
void calculate_need(int need[NUM_PROCESSES][NUM_RESOURCES], int allocated[NUM_PROCESSES][NUM_RESOURCES], int max_need[NUM_PROCESSES][NUM_RESOURCES]) {
    for (int i = 0; i < NUM_PROCESSES; i++) {
        for (int j = 0; j < NUM_RESOURCES; j++) {
            need[i][j] = max_need[i][j] - allocated[i][j];
        }
    }
}

// 判断一个进程是否需要的资源是否小于等于可用资源
bool check_process(int process, int need[NUM_PROCESSES][NUM_RESOURCES], int available[NUM_RESOURCES]) {
    for (int i = 0; i < NUM_RESOURCES; i++) {
        if (need[process][i] > available[i]) {
            return false;
        }
    }
    return true;
}

// 判断当前状态是否安全
bool check_safe(int available[NUM_RESOURCES], int allocated[NUM_PROCESSES][NUM_RESOURCES], int max_need[NUM_PROCESSES][NUM_RESOURCES]) {
    int need[NUM_PROCESSES][NUM_RESOURCES];
    calculate_need(need, allocated, max_need);
    bool finish[NUM_PROCESSES] = {false};
    int work[NUM_RESOURCES];
    for (int i = 0; i < NUM_RESOURCES; i++) {
        work[i] = available[i];
    }
    int safe_seq[NUM_PROCESSES];
    int count = 0;
    while (count < NUM_PROCESSES) {
        bool found = false;
        for (int i = 0; i < NUM_PROCESSES; i++) {
            if (!finish[i] && check_process(i, need, work)) {
                finish[i] = true;
                for (int j = 0; j < NUM_RESOURCES; j++) {
                    work[j] += allocated[i][j];
                }
                safe_seq[count] = i;
                count++;
                found = true;
            }
        }
        if (!found) {
            return false;
        }
    }
    return true;
}

// 请求资源
bool request_resources(int process, int request[NUM_RESOURCES], int available[NUM_RESOURCES], int allocated[NUM_PROCESSES][NUM_RESOURCES], int max_need[NUM_PROCESSES][NUM_RESOURCES]) {
    int need[NUM_PROCESSES][NUM_RESOURCES];
    calculate_need(need, allocated, max_need);
    for (int i = 0; i < NUM_RESOURCES; i++) {
        if (request[i] > need[process][i]) {
            return false;
        }
        if (request[i] > available[i]) {
            return false;
        }
    }
    for (int i = 0; i < NUM_RESOURCES; i++) {
        available[i] -= request[i];
        allocated[process][i] += request[i];
        need[process][i] -= request[i];
    }
    if (!check_safe(available, allocated, max_need)) {
        for (int i = 0; i < NUM_RESOURCES; i++) {
            available[i] += request[i];
            allocated[process][i] -= request[i];
            need[process][i] += request[i];
        }
        return false;
    }
    return true;
}

// 打印状态
void print_status() {
    printf("可用资源：");
    for (int i = 0; i < NUM_RESOURCES; i++) {
        printf(" %d", available[i]);
    }
    printf("\n最大需求：\n");
    for (int i = 0; i < NUM_PROCESSES; i++) {
        printf("进程 %d:", i);
        for (int j = 0; j < NUM_RESOURCES; j++) {
            printf(" %d", max_need[i][j]);
        }
        printf("\n");
    }
    printf("已分配资源：\n");
    for (int i = 0; i < NUM_PROCESSES; i++) {
        printf("进程 %d:", i);
        for (int j = 0; j < NUM_RESOURCES; j++) {
            printf(" %d", allocated[i][j]);
        }
        printf("\n");
    }
}

// 测试代码
int main() {
    print_status();

    if (check_safe(available, allocated, max_need)) {
        printf("当前状态安全\n");
    } else {
        printf("当前状态不安全\n");
    }

    int request[NUM_RESOURCES] = {1, 0, 2};
    if (request_resources(1, request, available, allocated, max_need)) {
        printf("请求资源成功\n");
        print_status();
    } else {
        printf("请求资源失败\n");
    }

    return 0;
}

运行上述代码，将输出以下结果：

可用资源： 3 3 2
最大需求：
进程 0: 7 5 3
进程 1: 3 2 2
进程 2: 9 0 2
进程 3: 2 2 2
进程 4: 4 3 3
已分配资源：
进程 0: 0 1 0
进程 1: 2 0 0
进程 2: 3 0 2
进程 3: 2 1 1
进程 4: 0 0 2
当前状态安全
请求资源成功
可用资源： 2 3 0
最大需求：
进程 0: 7 5 3
进程 1: 3 2 2
进程 2: 9 0 2
进程 3: 2 2 2
进程 4: 4 3 3
已分配资源：
进程 0: 0 1 0
进程 1: 3 0 2
进程 2: 3 0 2
进程 3: 2 1 1
进程 4: 0 0 2

可以看到，初始状态下当前状态是安全的，请求资源成功后仍然是安全状态。