操作系统代码实现：Number Project Name Content Summary State Type 一、Process Scheduling Algorithm Simulation 1、 Simulate the operation of the round-robin algorithm for process scheduling. 2、 Create at least 15 processes and output their scheduling situation under the scheduling algorithm mentioned above and output it to theterminal to check the execution of the algorithm. 3、 The output should include the arrival time of the processes, the end time, and the average execution time. Essential. General 二、Readers-Writer Problem Implmentation 1、 A data set is shared among several concurrent processes: Readers – only read the data set; they do not perform any updates. Writers – can both read and write. 2、 Problem – allow multiple readers (at most 8) to read at the same time. Only one single writer can access the shared data at the same time. Essential. General 三、Program for Least Recently used Algorithm 1、 Create a page access sequence (page number range 0-18) using a random function. The sequence length is 54 and assume that the number of main memory frames allocated to the thread is 6, that is, M = 6. 2、 Implement the LRU algorithm for page replacement on the above access sequence. 3、 Output the page replacement sequence and the page fault rate. Essential. General Requirements 1、 For each experiment project, submit a design report and code. The code should be implemented in C++. The requirements are as follows: a) The content of the design report should include the design ideas and implementation. b) The results of the design report should include testing and running results (screenshots of screen outputs). c) The conclusion of the design report should summarize the problems encountered, the solutions and experiences during the implementation process.

这是一个操作系统实现的项目，包含三个任务：模拟轮转算法进行进程调度、实现读者-写者问题、实现最近最少使用算法。第一项任务需要创建至少15个进程，并输出它们在轮转算法下的调度情况，包括到达时间、结束时间和平均执行时间等信息。第二项任务涉及到多个并发进程共享数据集，读者只读取数据集，而写者可以读写，需要实现允许多个读者同时读取，但只能有一个写者访问共享数据。第三项任务需要使用随机函数创建一个页面访问序列，长度为54，采用LRU算法进行页面替换，并输出页面替换序列和页面缺失率等信息。每个实验项目需要提交设计报告和代码，代码实现应为C++。设计报告的内容包括设计思路、实现细节、测试和运行结果等信息，同时需要总结实现过程中遇到的问题、解决方案和经验。

相关问题

请用C++完成操作系统代码（需要有注释）：一、Process Scheduling Algorithm Simulation 1、 Simulate the operation of the round-robin algorithm for process scheduling. 2、 Create at least 15 processes and output their scheduling situation under the scheduling algorithm mentioned above and output it to theterminal to check the execution of the algorithm. 3、 The output should include the arrival time of the processes, the end time, and the average execution time.

好的，我为您提供一份 C++ 代码，实现了轮转调度算法的进程调度模拟。代码中包括了注释，方便理解。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

// 进程类
class Process {
public:
    int pid;            // 进程id
    int arrival_time;   // 到达时间
    int service_time;   // 服务时间
    int start_time;     // 开始时间
    int end_time;       // 结束时间
    int remain_time;    // 剩余服务时间
};

int main() {
    const int time_slice = 2;   // 时间片大小
    vector<Process> process_list;   // 进程队列
    int total_turnaround_time = 0;  // 总周转时间

    // 创建15个进程
    for (int i = 0; i < 15; ++i) {
        Process p;
        p.pid = i+1;    // 进程id从1开始
        p.arrival_time = rand() % 10;   // 到达时间在[0,9]之间随机
        p.service_time = rand() % 10 + 1;   // 服务时间在[1,10]之间随机
        p.remain_time = p.service_time;
        process_list.push_back(p);
    }

    // 按到达时间从小到大排序
    sort(process_list.begin(), process_list.end(), [](const Process& p1, const Process& p2) {
        return p1.arrival_time < p2.arrival_time;
    });

    // 遍历进程队列，模拟轮转调度算法
    int current_time = 0;
    while (!process_list.empty()) {
        Process& p = process_list.front();  // 取出队首进程
        if (p.arrival_time > current_time) {    // 进程未到达
            current_time = p.arrival_time;
        }
        p.start_time = current_time;    // 记录进程开始时间
        int execute_time = min(time_slice, p.remain_time);  // 执行时间不超过时间片大小和剩余服务时间
        p.remain_time -= execute_time;  // 更新进程的剩余服务时间
        current_time += execute_time;   // 更新当前时间
        if (p.remain_time == 0) {       // 进程执行完毕
            p.end_time = current_time;  // 记录进程结束时间
            total_turnaround_time += p.end_time - p.arrival_time;   // 累加周转时间
            process_list.erase(process_list.begin());  // 将进程从队列中删除
        } else {                        // 进程未执行完毕，加入队尾
            process_list.push_back(p);
            process_list.erase(process_list.begin());  // 将进程从队首移到队尾
        }
    }

    // 输出进程的调度情况
    cout << "PID\tArrival Time\tEnd Time\tTurnaround Time" << endl;
    for (const Process& p : process_list) {
        cout << p.pid << "\t" << p.arrival_time << "\t\t" << p.end_time << "\t\t" << p.end_time - p.arrival_time << endl;
    }

    // 输出平均周转时间
    double avg_turnaround_time = (double)total_turnaround_time / 15;
    cout << "Average Turnaround Time: " << avg_turnaround_time << endl;

    return 0;
}

上述代码中使用了 C++11 中的 lambda 表达式，方便定义排序函数。程序输出的内容包括进程的 PID、到达时间、结束时间和周转时间，以及平均周转时间。

请用C++代码编写完成以下要求需要有注释：一、Process Scheduling Algorithm Simulation 1、 Simulate the operation of the round-robin algorithm for process scheduling. 2、 Create at least 15 processes and output their scheduling situation under the scheduling algorithm mentioned above and output it to theterminal to check the execution of the algorithm. 3、 The output should include the arrival time of the processes, the end time, and the average execution time.

好的，以下是注释后的C++代码实现：

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>

using namespace std;

const int maxn = 10010;

int n, q, cnt = 0, sum = 0;
int vis[maxn], ans[maxn], arr[maxn], endt[maxn], startt[maxn];
queue<int> qe;

struct Process {
    int id, arrive, left, rest;
} p[maxn];

bool cmp(Process a, Process b) {
    return a.arrive < b.arrive;
}

int main() {
    memset(vis, 0, sizeof(vis));
    memset(ans, 0, sizeof(ans));
    memset(arr, 0, sizeof(arr));
    memset(endt, 0, sizeof(endt));
    memset(startt, 0, sizeof(startt));
    //初始化

    cin >> n >> q;  //输入进程数和时间片大小

    for(int i = 0; i < n; i++) {
        cin >> p[i].arrive >> p[i].rest;
        p[i].id = i + 1;
        p[i].left = p[i].rest;
    }
    //输入进程的到达时间和执行时间，初始化进程的编号、剩余时间

    sort(p, p + n, cmp);    //按到达时间排序

    int now = p[0].arrive;  //当前时间初始化为第一个进程的到达时间
    qe.push(p[0].id);       //将第一个进程加入队列

    while(!qe.empty()) {
        int u = qe.front();
        qe.pop();

        if(!vis[u]) {   //进程第一次被调度
            startt[u] = max(now, arr[u]);  //开始时间为 当前时间 或 上一个进程结束时间
            endt[u] = startt[u] + min(q, p[u - 1].left);    //结束时间为 开始时间 + 时间片大小 或 进程运行剩余时间 
            p[u - 1].left -= min(q, p[u - 1].left);  //更新进程剩余时间
            now = endt[u];  //当前时间更新为进程结束时间

            if(p[u - 1].left > 0) qe.push(u);  //进程未运行完，重新加入队列
            else {
                vis[u] = 1; //进程已运行完
                cnt++;  //已运行完的进程数量加1
            }
        }
        else {  //进程已被调度过
            endt[u] = now + min(q, p[u - 1].left);    //更新结束时间
            p[u - 1].left -= min(q, p[u - 1].left);  //更新进程剩余时间
            now = endt[u];  //当前时间更新为进程结束时间

            if(p[u - 1].left > 0) qe.push(u);  //进程未运行完，重新加入队列
            else cnt++;    //进程已运行完
        }

        if(cnt == n) break; //所有进程都已运行完，跳出循环
        else {
            for(int i = 0; i < n; i++) {
                if(p[i].arrive <= now && !vis[p[i].id]) qe.push(p[i].id);   //将已到达未运行的进程加入队列
            }
        }
    }

    for(int i = 0; i < n; i++) {
        sum += endt[i + 1] - p[i].arrive;  //计算所有进程的总执行时间
    }

    double avg = (double)sum / n;   //计算平均执行时间

    cout << "Process\tArrive Time\tEnd Time\tAverage Time" << endl;
    for(int i = 0; i < n; i++) {
        printf("%d\t\t%d\t\t%d\t\t%d\n", p[i].id, p[i].arrive, endt[i + 1], endt[i + 1] - p[i].arrive);  //输出进程的编号、到达时间、结束时间、执行时间
    }
    printf("\nAverage Time: %.2lf\n", avg);  //输出平均执行时间

    return 0;
}

注：本代码实现的时间片为固定大小，如果要实现动态时间片可自行修改。