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CS 2110 Timed Lab 3: Subroutines and Calling Conventions solved

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2 Overview
2.1 Purpose
The purpose of this timed lab is to test your understanding of implementing subroutines in the LC-3 assembly
language using the calling convention, from both the callee and caller side.
2.2 Task
You will implement the subroutines listed below in LC-3 assembly language. Please see the detailed instructions for the subroutines on the following pages. We have provided pseudocode for the subroutines—you
should follow these algorithms when writing your assembly code. Your subroutines must adhere to the LC-3
calling conventions.
2.3 Criteria
Your assignment will be graded based on your ability to correctly translate the given pseudocode for a
subroutine (function) into LC-3 assembly code, following the LC-3 calling convention. Please use the LC-3
instruction set when writing these programs. Check the Deliverables section for what you must submit to
Gradescope.
You must produce the correct return values for each function. In addition, registers R0-R5 and R7 must be
restored from the perspective of the caller, so they contain the same values after the caller’s JSR subroutine
call. Your subroutine must return to the correct point in the caller’s code, and the caller must find the
return value on the stack where it is expected to be. If you follow the LC-3 calling conventions correctly, all
of these things will happen automatically. Additionally, we will check that you made the correct subroutine
calls, so you should not try to implement a recursively subroutine iteratively.
Your code must assemble with no warnings or errors (Complx and the autograder will tell you if there are
any). If your code does not assemble, we will not be able to grade that file and you will not receive any
points.
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3 Detailed Instructions
For this Timed Lab, you will be implementing three subroutines: MAX, DIV and MAP. MAX is a non-recursive
subroutine that will return 0 if the first argument is larger than the second, and 1 if the second argument is
at least as large as the first. DIV is a recursive subroutine that will divide the first argument by the second.
MAP will iterate through an array, one pair of elements at a time, and replace the larger of the two elements
with the first element divided by the second element.
3.1 MAX subroutine
MAX will take two arguments a and b and return 0 if a is greater than b, and 1 if b is greater than or equal
to a.
Note: If a and b are equal, you should return 1 as demonstrated in the pseudocode.
You should implement the subroutine as shown in the pseudocode below:
// checkpoint 1
int MAX(int a, int b) {
if (a > b) {
return 0;
} else {
return 1;
}
}
Examples:
• MAX(2, 5) returns 1
• MAX(1, -4) returns 0
• MAX(3, 3) returns 1
Implementing MAX will be the first checkpoint. Please refer to Checkpoints for details on how MAX: will be
graded.
3.2 DIV subroutine
DIV will take two non-negative integer arguments x and y (y will also be non-zero) and compute x/y:
that is, x divided by y.
Note: This is integer division, the resulting quotient is truncated. You should implement the subroutine
as shown in the pseudocode below:
DIV(x, y) {
// (checkpoint 2) – Base Case
if (y > x) {
return 0;
// (checkpoint 3) – Recursion
} else {
return 1 + DIV(x – y, y);
}
}
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Examples:
• DIV(9, 3) returns 3
• DIV(3, 5) returns 0
• DIV(26, 5) returns 5
Note: All of the arguments for this function are guaranteed to be non-negative. y is also guaranteed to
be non-zero.
DIV contains two checkpoints. Checkpoint 2 is the base case: returning 0 when y > x. Checkpoint 3 is the
recursive case: DIV should return the correct value for all non-negative integers where y > 0. Please refer
to Checkpoints for details on how DIV will be graded.
3.3 MAP subroutine
MAP takes two arguments:
• array: The address of the array to be modified
• length: The length of the array to be modified
Note: The argument array is guaranteed to have an even length. All the elements of MAP are guaranteed
to be non-negative.
MAP will iterate through the array one pair of consecutive elements at a time. For each pair of elements, MAP
will apply DIV to divide the first element by the second element and replace the larger of the two elements
with the result. MAP will determine the index of the larger element by applying MAX to the pair and adding
the result to the first element’s index.
Note: The pairs that MAP iterates through should not overlap, meaning the first pair would be the elements
at indices 0 and 1, the second pair would be the elements at indices 2 and 3, etc.
You should implement the subroutine as shown in the pseudocode below:
// (checkpoint 4)
void MAP(array, length) {
int i = 0;
while (i < length) { int firstElem = arr[i]; int secondElem = arr[i + 1]; int div = DIV(firstElem, secondElem); int offset = MAX(firstElem, secondElem); arr[i + offset] = div; i += 2; } } Examples: • Before: array = [12, 5]; length = 2 – After: array = [2, 5] • Before: array = [8, 2, 0, 7, 12, 1, 5, 5]; length = 8 4 – After: array = [4, 2, 0, 0, 12, 1, 5, 1] You do not need to return a value from MAP, you should instead modify array directly. While you will still leave a spot for a return value on the stack, it will be ignored. Correctly implementing MAP is checkpoint 4. Please refer to Checkpoints for details on how MAP will be graded. 5 4 Checkpoints 4.1 Checkpoints (70 points) In order to get all of the points for this timed lab, your code must meet these checkpoints: • Checkpoint 1 (15 points): In MAX, load and compare the parameters a and b to return 0 when a > b
and 1 when b >= a.
• Checkpoint 2 (15 points): Implement the base case for DIV to return 0 when y > x.
• Checkpoint 3 (20 points): Implement the recursive case of DIV to successfully compute and return x/y
for any non-negative integer x and any positive integer y.
• Checkpoint 4 (20 points): Implement MAP. You should iterate through the array one pair of elements
at a time, and use the MAX subroutine to determine the index of the larger element of the pair. You
should then apply the DIV subroutine to the pair. The results of each DIV subroutine call must be
properly stored back into the array at the index of the larger element.
4.2 Other Requirements (30 points)
Your subroutine must follow the LC-3 calling convention. Specifically, it must fulfill the following conditions:
• Your DIV subroutine must be recursive and call itself according to the pseudocode’s description.
• When your subroutine returns, every register must have its original value preserved (except R6).
• When your subroutine returns, the stack pointer (R6) must be decreased by 1 from its original value
so that it now points to the return value.
• During the execution of your subroutine, you must make the correct number of calls to MAX and DIV,
corresponding to the pseudocode.
– If the autograder claims that you are making an unknown subroutine call to some label in your
code, it may be that your code has two labels without an instruction between them. Removing
one of the labels should appease the autograder.
5 Deliverables
Turn in the following files on Gradescope during your assigned timed lab slot:
1. tl03.asm
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6 Local Autograder
To run the autograder locally, follow the steps below depending upon your operating system:
• Mac/Linux Users:
1. Navigate to the directory your homework is in (in your terminal on your host machine, not
in the Docker container via your browser)
2. Run the command sudo chmod +x grade.sh
3. Now run ./grade.sh
• Windows Users:
1. In Git Bash (or Docker Quickstart Terminal for legacy Docker installations), navigate to the
directory your homework is in
2. Run chmod +x grade.sh
3. Run ./grade.sh
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7 LC-3 Assembly Programming Requirements
7.1 Overview
1. Your code must assemble with NO WARNINGS OR ERRORS. To assemble your program, open
the file with Complx. It will complain if there are any issues. If your code does not assemble, you
WILL get a zero for that file.
2. Comment your code! This is especially important in assembly, because it’s much harder to interpret
what is happening later, and you’ll be glad you left yourself notes on what certain instructions are
contributing to the code. Comment things like what registers are being used for and what less intuitive
lines of code are actually doing. To comment code in LC-3 assembly just type a semicolon (;), and the
rest of that line will be a comment.
3. Avoid stating the obvious in your comments, it doesn’t help in understanding what the code is doing.
Good Comment
ADD R3, R3, -1 ; counter–
BRp LOOP ; if counter == 0 don’t loop again
Bad Comment
ADD R3, R3, -1 ; Decrement R3
BRp LOOP ; Branch to LOOP if positive
4. DO NOT assume that ANYTHING in the LC-3 is already zero. Treat the machine as if your
program was loaded into a machine with random values stored in the memory and register file.
5. Following from 4., you can randomize the memory and load your program by going to File ¿ Advanced
Load and selecting RANDOMIZE for registers and memory.
6. Use the LC-3 calling convention. This means that all local variables, frame pointer, etc., must be
pushed onto the stack. Our autograder will be checking for correct stack setup.
7. The stack will start at xF000. The stack pointer always points to the last used stack location.
This means you will allocate space first, then store onto the stack pointer.
8. Do NOT execute any data as if it were an instruction (meaning you should put HALT or RET instructions
before any .fills).
9. Do not add any comments beginning with @plugin or change any comments of this kind.
10. You should not use a compiler that outputs LC3 to do this assignment.
11. Test your assembly. Don’t just assume it works and turn it in.
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8 Appendix
8.1 Appendix A: LC-3 Instruction Set Architecture
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