合肥生活安徽新聞合肥交通合肥房產(chǎn)生活服務(wù)合肥教育合肥招聘合肥旅游文化藝術(shù)合肥美食合肥地圖合肥社保合肥醫(yī)院企業(yè)服務(wù)合肥法律

        代做CITS2002、C/C++語言程序代寫
        代做CITS2002、C/C++語言程序代寫

        時間:2024-10-09  來源:合肥網(wǎng)hfw.cc  作者:hfw.cc 我要糾錯



        CITS2002 - Second Project
        A simple simulation of virtual memory
        • This project is worth 10% of the marks in the unit.
        • The project can be done in groups of two.
        • The due date of the project is October 17, 11:59 pm.
        • The project description is long, but the coding is simple. We will
        discuss the project in the workshops on Fridays.
        1 A simple simulation of virtual memory
        The aim of this project is to simulate a simple virtual memory system using
        an array as the RAM of a hypothetical machine. The project will also require
        some C programming skills of using structures and pointers.
        We have a computer whose RAM is an array of size 16. It is an array
        of pointers. There are 8 page frames in the RAM, each consisting of two
        contiguous locations in the array. Hence, the page size of this computer is 2.
        The virtual memory of this computer is an array of pointers of size **
        (We will pretend it is on disc, but actually it is an array in the RAM of
        our computer). There are 4 processes in this computer, and each proces can
        have 4 pages, and obviously all the pages of all the processes cannot be in
        the main memory at the same time. Some pages will be in the main memory
        and some pages will be in the virtual memory at any time. The processes are
        numbered 0 . . . 3. Each process has a page table, which is an integer array,
        entry of a process page table indicates whether the page is in RAM or in the
        virtual memory (on disc), k if the page is in RAM (k is the frame number,
        between 0 . . . 7), and 99 if the page is in disc (99 cannot be a frame number).
        You have to define a structure that will consist of three fields, a process
        id, a page number of the process, and the last time this page was accessed
        if it is in the RAM. Time in the simulation is not real time, rather a time
        step. Time increases in simulation steps, as explained below. The simulation
        starts (at time 0) by initializing the virtual memory with all the 4 pages of
        each process. You have to do the following steps before the simulation starts:
        1• Define a structure whose pointer will be stored in each array location
        of the RAM and the virtual memory. The structure may look like this:
        struct {
        int process_id;
        int page_num;
        int last_accessed;
        } memory;
        Initialise the process id and page num with the id of the process (a
        number between 0 . . . 3) and a page number of that process (a number
        between 0 . . . 3). Initialise all last access to 0.
        • Create each page and store pointers in the array for the virtual memory.
         Note that the process id and page num of two consecutive array
        locations will be the same since each page occupies two array locations.
        The simulation starts by reading a file where there is a single line of
        integers separated by blanks, for example:
        0 2 1 3 3 2 2 0 2 1 0 2 3 0
        Each integer indicates a process id. For example, the first number 0 indicates
         that the next page of process 0 has to be brought in from virtual
        memory to the RAM. The process table of process 0 and the RAM have to
        be updated accordingly. You can keep the content of the virtual memory
        unchanged, as that is how virtual memory systems work. Our processes do
        not do any computation, they just request the next page and later may write
        a page back to virtual memory. You can assume for simplicity that all the
        pages are always in the virtual memory and nothing needs to be written
        back, as no page is updated by doing any computation. The last accessed
        time of a page will be the time step when you brought the page to RAM.
        For example, after reading this file, the first (or 0th page of process 0 will
        be brought to RAM), the last accessed time of this page will be 0, as the
        simulation starts now and time is 0. Time will increase by 1 for each entry
        in the file.
        The RAM may become full sometime, you have to use the local Least
        Recently Used (LRU) algorithm for evicting a page and bringing a new page.
        2local means you have to evict the least recently used page of the same
        process for accommodating the new page. If there is no page of the process
        whose page you want to bring in, use a global LRU policy, evict the page
        that is least recently used among all pages in the RAM.
        2 Submission
        You have to write a C program in a single file called simulation.c, and
        compiled as an executable called simulation. It will read two file names from
        the command line, in.txt and out.txt. The first file is the one mentioned
        above, for reading process ids. The second file is an output file where you
        should print the following information at the end of the simulation. Your
        submission will be executed as:
        simulation in.txt out.txt
        • The page tables of the four processes in separate lines. For example,
        the page table for process 0 may look like this:
        3, 2, 1, 99
        This means there are three pages of process 0 in the RAM, pages 0, 1
        and 2, in frames 3, 2 and 1, and page 3 is in the disc.
        You have to also print the content of the RAM, each location separated
        by a ’;’. For example, the RAM may look like this:
        0,0,5; 0,0,5; 2,0,1; 2,0,1; etc. (16 entries)
        Note that, the first two locations of the RAM stores page 0 of process
        0, as each page occupies two array locations of the RAM. Also, this
        page was brought to RAM at time step 5.
        Amitava Datta
        September 2024


        請加QQ:99515681  郵箱:99515681@qq.com   WX:codinghelp








         

        掃一掃在手機打開當(dāng)前頁
      1. 上一篇:代做00301J 、Java設(shè)計編程代寫
      2. 下一篇: 代寫COMP612、代做C++程序設(shè)計
      3. 無相關(guān)信息
        合肥生活資訊

        合肥圖文信息
        急尋熱仿真分析?代做熱仿真服務(wù)+熱設(shè)計優(yōu)化
        急尋熱仿真分析?代做熱仿真服務(wù)+熱設(shè)計優(yōu)化
        出評 開團工具
        出評 開團工具
        挖掘機濾芯提升發(fā)動機性能
        挖掘機濾芯提升發(fā)動機性能
        海信羅馬假日洗衣機亮相AWE  復(fù)古美學(xué)與現(xiàn)代科技完美結(jié)合
        海信羅馬假日洗衣機亮相AWE 復(fù)古美學(xué)與現(xiàn)代
        合肥機場巴士4號線
        合肥機場巴士4號線
        合肥機場巴士3號線
        合肥機場巴士3號線
        合肥機場巴士2號線
        合肥機場巴士2號線
        合肥機場巴士1號線
        合肥機場巴士1號線
      4. 短信驗證碼 酒店vi設(shè)計 NBA直播 幣安下載

        關(guān)于我們 | 打賞支持 | 廣告服務(wù) | 聯(lián)系我們 | 網(wǎng)站地圖 | 免責(zé)聲明 | 幫助中心 | 友情鏈接 |

        Copyright © 2025 hfw.cc Inc. All Rights Reserved. 合肥網(wǎng) 版權(quán)所有
        ICP備06013414號-3 公安備 42010502001045

        日韩丰满少妇无码内射| 久久蜜桃精品一区二区三区| 中文成人无码精品久久久不卡| 久久亚洲日韩看片无码| 国产精品视频免费一区二区三区| 无码人妻精品一区二区三区蜜桃 | 国产伦精品一区二区三区不卡| 91久久精品国产免费一区| 亚洲AV乱码久久精品蜜桃| 国语自产少妇精品视频蜜桃| 996久久国产精品线观看| 精品多毛少妇人妻AV免费久久| 人妻少妇精品久久| 日韩视频中文字幕| 日韩好片一区二区在线看| 一本色道久久88综合日韩精品| 国产真实乱人偷精品| 国产精品午夜无码av体验区| 最新国产精品视频| 精品国产_亚洲人成在线| WWW夜片内射视频日韩精品成人| 思思久久96热在精品国产| 亚洲国产精品无码中文lv| 亚洲精品国产suv一区88| 国产精品久久久久久久app | 国产成人精品男人免费| 国产精品亚洲一区二区三区| 国产精品香蕉在线观看不卡| 国产精品资源在线| 国产成人综合日韩精品无码| 国产精品videossex另类| 国产精品久久久久久久网站| 国产精品美女一区二区| 国产精品va一区二区三区| 无码欧精品亚洲日韩一区夜夜嗨| 亚洲日韩精品一区二区三区无码| 日韩制服国产精品一区| 亚洲AV日韩AV一区二区三曲 | 成品人和精品人的区别在哪里| 99久久精品国产高清一区二区 | 亚洲中文字幕久久精品蜜桃|