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CS 61 - Programming Assignment 3
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)
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 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 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
● 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!
for(i = 15 downto
if (msb is a
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
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)
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
○ 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.
Left Shifted, with 0 shifted in to LSB:
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).
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.
● 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.