Assembly language lc-3

Computer Science


Question Description

Using these rigister write the code the instruction are provided in pdf

.ORIG x3000; Program begins here




LD R6, Value_ptr; R6 <-- pointer to value to be displayed as binary

LDR R1, R6, #0; R1 <-- value to be displayed as binary








Value_ptr .FILL xCB00 ; The address where value to be displayed is stored

.ORIG xCB00; Remote data

Value .FILL xABCD; <----!!!NUMBER TO BE DISPLAYED AS BINARY!!! Note: label is redundant.





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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 in a register, and display it to the console as 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 [#-32768, #32767] (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 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 the most significant bit, aka ​bit 15​)​ ​to right​ ​(the trailing bit, aka the rightmost bit, aka the least significant bit, aka ​bit 0​)​. 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 ​BR​anch instruction has parameters (n, z, p) which tell it to check whether the LMR (Last Modified Register) is ​n​egative, ​z​ero, or ​p​ositive (or any combination thereof). 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 h ​ ere​ . ● 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 if (msb is a print a else: print a shift left 0): 1): 1 0 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 N ​ O 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 n ​ o​ space character at the 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 ■ ● 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 When 1010 1010 ; #5 Left Shifted, with 0 shifted in to LSB: <---- 0101 ; #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 ● 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 grade will 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. Comics??! Sweet!! Source: ...
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