<samp id="e4iaa"><tbody id="e4iaa"></tbody></samp>
<ul id="e4iaa"></ul>
<blockquote id="e4iaa"><tfoot id="e4iaa"></tfoot></blockquote>
    • <samp id="e4iaa"><tbody id="e4iaa"></tbody></samp>
      <ul id="e4iaa"></ul>
      <samp id="e4iaa"><tbody id="e4iaa"></tbody></samp><ul id="e4iaa"></ul>
      <ul id="e4iaa"></ul>
      <th id="e4iaa"><menu id="e4iaa"></menu></th>

      CSC 256代寫、C++設計編程代做

      時間:2024-04-30  來源:  作者: 我要糾錯



      CSC 256-04 Final Project Part 2 (50 Points)
      An ancient robot game
      For this project, you are given a program that implements a game where a human user
      tries to escape from four robots. The user and four robots are on an x-y grid. On every
      step, you enter a move for the human. The robots will attempt to get closer to the human.
      When a robot has the same x-y coordinates as the human, the game is over. You will
      translate this program faithfully, following all function call guidelines and MIPS register
      use conventions.
      Two arrays x[4] and y[4] keep track of the x- and y-coordinates of four robots. The
      positions of the human and the four robots are initialized in the program. On each step,
      the user enters a move; the positions of the human and the robots are updated. This
      continues until the human dies. This figure gives an idea of the game; we’ll work with a
      text-only version, sorry.
      In the main loop, the user is prompted to enter a move. The position of the user is
      updated. Then the program calls a function moveRobots() to update the position of
      the robots as they try to catch the human. The new positions of the human and the robots
      are then displayed.
      The function moveRobots() has prototype
      int moveRobots(int *arg0, int *arg1, int arg2, int arg3)
      arg0 is the base address of array that contains the x-coordinates of the four robots, arg1 is
      the base address of array that contains the y-coordinates of the four robots, arg2 is the
      x-coordinate of the human, arg3 is the y-coordinate of the human.
      moveRobots()updates the positions of the four robots, and returns a 1 if the human is
      alive, and a 0 if the human is dead (i.e., the human has the same coordinates as a robot).
      Each coordinate of a robot is updated by calling the function getNew(), which returns
      the new coordinate based on the current coordinate of the robot and the current
      coordinate of the human.
      When you translate moveRobots() to MIPS assembly language, arg0 through arg3
      are in $a0 through $a3; the return value is in $v0.
      The function getNew() uses simple rules to move a robot closer to the human. If the
      difference in the coordinates is >=10, the robot's coordinate will move 10 units closer to
      the human. If the difference in the coordinates is < 10, the robot's coordinate will
      move one unit closer to the human. (See program listings.) getNew() has prototype
      int getNew(int arg0, int arg1)
      arg0 is the coordinate (x or y) of a robot, arg1 is the coordinate (x or y) of the human.
      getNew() returns the new coordinate of the robot, based on the position of the human.
      The function getNew() is already translated to MIPS assembly language, arg0 and arg1
      are in $a0 and $a1 respectively, and the return value is in $v0.
      A copy of the C++ program robots.cpp can be found here. The file assembly file contains
      the main program and getnew(), already translated into MIPS assembly language. Your
      functions will follow the main program in the same file. The template assembly file can
      be found on Canvas.
      Write the functions exactly as described in this handout. Do not implement the
      program using other algorithms or tricks. Do not even switch the order of the arguments
      in function calls; you must follow the order specified in the C++ code. The purpose of
      this program is to test whether you understand nested functions. If you wish to make
      changes to the algorithm, you must first check with the instructor.
      Your functions should be properly commented. Each function must have its own header
      block, including the prototype of the function, the locations of all arguments and return
      values, descriptions of the arguments and how they are passed, and a description of what
      the function does. Paste in the C++ code as inline comments for your MIPS assembly
      code.
      Submission: submit your code via Canvas. All your code should be in a single plain text
      file(PDF only).
      80% of your grade is for correctness. 20% is for clarity/documentation.
      Output%
      Your coordinates: 25 25
      Enter move (1 for +x, -1 for -x, 2 for + y, -2 for -y):2
      Your coordinates: 25 26
      Robot at 10 10
      Robot at 10 40
      Robot at 40 10
      Robot at 40 40
      Enter move (1 for +x, -1 for -x, 2 for + y, -2 for -y):1
      Your coordinates: 26 26
      Robot at 20 20
      Robot at 20 30
      Robot at 30 20
      Robot at 30 30
      Enter move (1 for +x, -1 for -x, 2 for + y, -2 for -y):2
      Your coordinates: 26 27
      Robot at 21 21
      Robot at 21 29
      Robot at 29 21
      Robot at 29 29
      Enter move (1 for +x, -1 for -x, 2 for + y, -2 for -y):1
      Your coordinates: 27 27
      Robot at 22 22
      Robot at 22 28
      Robot at 28 22
      Robot at 28 28
      Enter move (1 for +x, -1 for -x, 2 for + y, -2 for -y):2
      Your coordinates: 27 28
      Robot at 23 23
      Robot at 23 28
      Robot at 27 23
      Robot at 27 28
      AAAARRRRGHHHHH... Game over
      libra%

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


















       

      標簽:

      掃一掃在手機打開當前頁
    • 上一篇:代做FINM7008、代寫FINM7008 Applied Investments
    • 下一篇:COMP30024代做、代寫 Game of Tetress
    • 無相關信息
      昆明生活資訊

      昆明圖文信息
      蝴蝶泉(4A)-大理旅游
      蝴蝶泉(4A)-大理旅游
      油炸竹蟲
      油炸竹蟲
      酸筍煮魚(雞)
      酸筍煮魚(雞)
      竹筒飯
      竹筒飯
      香茅草烤魚
      香茅草烤魚
      檸檬烤魚
      檸檬烤魚
      昆明西山國家級風景名勝區
      昆明西山國家級風景名勝區
      昆明旅游索道攻略
      昆明旅游索道攻略
    • NBA直播 短信驗證碼平臺 幣安官網下載 歐冠直播 WPS下載

      關于我們 | 打賞支持 | 廣告服務 | 聯系我們 | 網站地圖 | 免責聲明 | 幫助中心 | 友情鏈接 |

      Copyright © 2025 kmw.cc Inc. All Rights Reserved. 昆明網 版權所有
      ICP備06013414號-3 公安備 42010502001045

      主站蜘蛛池模板: 无码专区狠狠躁躁天天躁| 成人无码精品一区二区三区| 国产乱子伦精品无码专区 | 国产a v无码专区亚洲av| 精品人体无码一区二区三区| 久久午夜无码鲁丝片秋霞| 亚洲aⅴ天堂av天堂无码麻豆| 五十路熟妇高熟无码视频| 无码精品人妻一区二区三区人妻斩 | 亚洲av无码一区二区三区在线播放 | 精品欧洲av无码一区二区三区| 亚洲中文字幕无码专区| 亚洲日韩一区二区一无码| 亚洲人成网亚洲欧洲无码久久| 亚洲综合无码一区二区痴汉| 亚洲AV综合色区无码一区爱AV| 国产成人无码免费视频97| 一区二区三区无码被窝影院 | 精品无码久久久久久尤物| 亚洲Av无码国产情品久久| 无码高潮少妇毛多水多水免费| 精品人妻无码区在线视频| 亚洲国产精品无码久久一线| 无码人妻一区二区三区免费视频| 亚洲av永久无码精品网址| 直接看的成人无码视频网站| 中文字幕人妻无码系列第三区 | 久久无码av三级| 自慰无码一区二区三区| 国产无码一区二区在线| MM1313亚洲精品无码久久| 熟妇人妻无码xxx视频| 亚洲精品无码成人| 亚洲乱人伦中文字幕无码| 亚洲AV无码一区二区大桥未久 | 无码中文字幕日韩专区视频| 亚洲AV中文无码字幕色三| 久久久无码精品亚洲日韩蜜桃| 国产v亚洲v天堂无码网站| 久久久久亚洲AV无码麻豆| 日韩网红少妇无码视频香港|