合肥生活安徽新聞合肥交通合肥房產生活服務合肥教育合肥招聘合肥旅游文化藝術合肥美食合肥地圖合肥社保合肥醫院企業服務合肥法律

        CS 61程序代做、代寫C/C++編程設計

        時間:2024-02-08  來源:合肥網hfw.cc  作者:hfw.cc 我要糾錯



        CS 61 - Programming Assignment 3
        Objective
        The purpose of this assignment is to give you more practice with I/O, and with left-shifting, aka
        multiplying by 2, and useful 2’s complement logic.
        High Level Description
        Store a number to the memory address specified in your assn 3 template. In your program, load
        that number to a register, and display it to the console as a 16-bit two's complement binary (i.e.
        display the binary value stored in the register, as a sequence of 16 ascii '1' and '0' characters).
        Note: Valid numbers are [#-**768, #**767] (decimal) or [x0000, xFFFF] (hex)
        Your Tasks
        You do not yet know how to take a multi-digit decimal number from user ascii input and convert it
        to binary, so for this assignment you are going to let the assembler do that part for you:
        you will use the .FILL pseudo-op to take a literal (decimal or hex, as you wish) and translate it into a
        16-bit two's complement binary, which will be stored in the indicated memory location; and then
        you will Load that value from memory into a register.
        You MUST use the provided assn3.asm template to set this up: it ensures that the number to be
        converted is always stored in the same location (the memory address specified in your template)
        so we can test your work; make sure you fully understand the code fragment we provide.
        At this point, your value will be stored in, say, R1: it is now your job to identify the 1’s and 0’s from
        the number and print them out to the console one by one, from left (the leading bit, aka the
        leftmost bit, aka bit 15, aka the most significant bit) to right (the trailing bit, aka the rightmost bit,
        aka bit 0, aka the least significant bit).
        Important things to consider:
        ● Recall the difference between a positive number and a negative number in 2’s complement
        binary: if the most significant bit (MSB) is 0, the number is considered positive (or zero);
        if it is 1, the number is negative.
        ● The BRanch instruction has parameters (n, z, p) which tell it to check whether the LMR (Last
        Modified Register) is negative, zero, or positive (or any combination of such tests).
        Hint: what can you say about the msb of the LMR if the n branch is taken?
        Review the workings of the NZP condition codes and the BR instruction here .
        ● Once you are done inspecting the MSB and printing the corresponding ascii '0' or '1', how would
        you shift the next bit into its place so you could perform the next iteration?
        Hint: the answer is in the objectives!
        Pseudocode:
        for(i = 15 downto 0):
        if (msb is a 1):
        print a 1
        else:
        print a 0
        shift left
        Note on creating LC-3 "control structures"
        See here for tips on creating LC-3 versions of the branch and loop control structures you are familiar
        with from C++ (Resources -> LC-3 Resources -> LC3 Assembly Language -> Control Structures in LC3)
        Expected/ Sample output
        In this assignment, your output will simply be the contents of R1, printed out as 16 ascii 1's and 0's,
        grouped into packets of 4, separated by spaces (as always, newline terminated, but with NO
        terminating space!)
        So if the hard coded value was xABCD, your output will be:
        1010 1011 1100 1101
        (The value stored to memory with .FILL was xABCD)
        Note:
        1. There are spaces after the first three "packets" of 4 bits (but no space character at end!)
        2. There is a newline after the output - again, there is NO space before the newline
        3. You must use the memory address specified in your template to hold the value to be output
        Your code will obviously be tested with a range of different values: Make sure you test your code
        likewise!
        Uh…help?
        ● MSB
        ○ Stands for Most Significant Bit
        ■ aka “left most bit” or “leading bit” or bit 15
        ○ When MSB is 0:
        ■ Means that the number is Not Negative (Positive or Zero)
        ○ When MSB is 1:
        ■ Means that the number is Negative
        ○ Further Reading
        ■ https://en.wikipedia.org/wiki/Most_significant_bit
        ● Left Shifting
        Left shifting means that you shift all the bits to the left by 1: so the MSB is lost, and is replaced
        by the bit on its right. A 0 is "shifted in" on the right to replace the previous LSB.
        4-bit Example:
        0101 ; #5
        When Left Shifted, with 0 shifted in to LSB:
        1010 <---- 0101
        1010 ; #10
        What happened when we left shifted? How did the number change?
        When left shifting, the number gets multiplied by 2? Why 2?
        Well, what happens when you shift a decimal number one place to the left? Why?
        (Practical differences between decimal and binary numbers are that we don't usually limit
        decimal numbers to a specific number of places, nor do we usually pad them with leading zeros).
        Further Reading
        ● https://en.wikipedia.org/wiki/Logical_shift
        Submission Instructions
        Submit ("Upload") your assignment3.asm file (and ONLY that file!) to the Programming Assignment 3
        folder in Gradescope: the Autograder will run & report your grade within a minute or so.
        You may submit as many times as you like - your assignment grade will normally be that of your last
        submission.
        If you wish to set your grade to a previous submission with a higher score, you may open your
        "Submission history" and "Activate" any other submission - that's the one we will see.
        Rubric
        ● To pass the assignment, you need a score of >= 80%.
        The autograder will run several tests on your code, and assign a grade for each.
        But certain errors (run-time errors, incorrect usage of I/O routines, missing newlines, etc.) may
        cause ALL tests to fail => 0/100! So submit early and study the autograder report carefully!!
        ● You must use the template we provide - if you make any changes to the provided starter code,
        the autograder may not be able to interpret the output, resulting in a grade of 0.
        如有需要,請加QQ:99515681 或WX:codehelp

        掃一掃在手機打開當前頁
      1. 上一篇:CISC3025代做、代寫Java,c++設計編程
      2. 下一篇:代寫Computer Security and Networks編程
      3. 無相關信息
        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+熱設計優化
        急尋熱仿真分析?代做熱仿真服務+熱設計優化
        出評 開團工具
        出評 開團工具
        挖掘機濾芯提升發動機性能
        挖掘機濾芯提升發動機性能
        海信羅馬假日洗衣機亮相AWE  復古美學與現代科技完美結合
        海信羅馬假日洗衣機亮相AWE 復古美學與現代
        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設計 NBA直播 幣安下載

        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 hfw.cc Inc. All Rights Reserved. 合肥網 版權所有
        ICP備06013414號-3 公安備 42010502001045

        主站蜘蛛池模板: 精品无码国产AV一区二区三区| 中文字幕人妻第一区| 日韩国产精品无码一区二区三区| 国产一区二区三精品久久久无广告| 无码国产精成人午夜视频一区二区| 国产在线精品一区二区| 伊人激情AV一区二区三区| 日本一区二区三区在线观看视频| 久久久久久综合一区中文字幕| 国产萌白酱在线一区二区| 国产乱码精品一区二区三区四川人| 久久精品无码一区二区无码 | 日韩爆乳一区二区无码| 日韩精品一区二区三区不卡| 国精产品一区一区三区MBA下载| 无码人妻少妇色欲AV一区二区| 日韩人妻无码一区二区三区 | 怡红院一区二区三区| 无码乱人伦一区二区亚洲 | 无码人妻aⅴ一区二区三区有奶水| 无码少妇精品一区二区免费动态| 精品视频一区二区三区四区五区| 日韩在线一区高清在线| 精品国产一区二区三区免费看| 亚洲AV成人一区二区三区观看 | 97精品国产一区二区三区| 精品无码人妻一区二区免费蜜桃 | 国产情侣一区二区三区| 一区三区三区不卡| 狠狠做深爱婷婷久久综合一区| 91国偷自产一区二区三区| 国产精品一级香蕉一区| 亚洲一区二区三区在线观看蜜桃| 色狠狠色噜噜Av天堂一区| 国内精品无码一区二区三区| 国产一区二区精品| 91秒拍国产福利一区| 麻豆天美国产一区在线播放| 午夜影院一区二区| 亚洲色无码专区一区| 无码一区二区三区在线|