ICO Lab 03 — Answers and Detailed Solutions

Introduction to Computer Organization (CCIT4026)
Lab 03: Calculator (Addition, Subtraction, Multiplication, Division, Logical Shift)

This document gives answers to the lab tasks and detailed explanations for each part of the solution.

1. Understanding the Problem

1.1 Formula

The program must compute:

Result = (quotient : remainder) = ((B + C) × A − (D ≪ 4)) / E

Where: - (D ≪ 4) means D left-shifted by 4 bits (same as D × 16). - quotient = integer division: var3 / E - remainder = modulo: var3 % E

1.2 Constraint on Inputs

The five integers must satisfy:

5 ≤ E < D < C < B < A ≤ 20

So: - E is the smallest (at least 5). - A is the largest (at most 20). - Order: E < D < C < B < A.

The lab asks the user to enter values in that order: E, then D, then C, then B, then A.

1.3 Variables to Compute and Store (in data memory)

Variable	Formula	Meaning
var1	B + C	First addition
var2	var1 × A	Multiplication (use lower 32 bits only)
var3	var2 − (D ≪ 4)	Subtract (D left-shifted by 4) from var2
quotient	var3 / E	Integer quotient of var3 ÷ E
remainder	var3 % E	Remainder of var3 ÷ E

2. Manual Calculation (Sample)

Use values that satisfy 5 ≤ E < D < C < B < A ≤ 20. Example: E=5, D=6, C=7, B=8, A=9.

Step	Calculation	Result
var1	B + C = 8 + 7	15
var2	var1 × A = 15 × 9	135
D ≪ 4	6 × 16	96
var3	var2 − (D ≪ 4) = 135 − 96	39
quotient	var3 / E = 39 / 5	7
remainder	var3 % E = 39 % 5	4

Note: In the lab PDF sample it says “var3 = var2 − (A ≪ 4)”. That is a typo; the formula uses (D ≪ 4), not (A ≪ 4). With D=6, (D ≪ 4) = 96, and 135 − 96 = 39.

3. MIPS Solution — Complete Assembly Program

The following program follows the lab’s pseudo-code steps and uses only instructions from the course (MARS syscalls 1, 4, 5, 10; add, sub, mul, div, mflo, mfhi, sll, lw, sw, li, la, jr, etc.).

# ICO Lab 03 - Calculator
# Formula: Result = (quotient : remainder) = ((B+C)*A - (D<<4)) / E
# Constraint: 5 <= E < D < C < B < A <= 20
# Input order: E, D, C, B, A

        .data
promptE:    .asciiz "Please enter the value E: "
promptD:    .asciiz "Please enter the value D: "
promptC:    .asciiz "Please enter the value C: "
promptB:    .asciiz "Please enter the value B: "
promptA:    .asciiz "Please enter the value A: "
msg_var1:   .asciiz "var1 = (B + C) = "
msg_var2:   .asciiz "var2 = var1 * A = "
msg_var3:   .asciiz "var3 = var2 - (D << 4) = "
msg_quot:   .asciiz "quotient = var3 / E = "
msg_rem:    .asciiz "remainder = var3 % E = "
newline:    .asciiz "\n"
author:     .asciiz "[Your Name, Class, Student ID]\n"

E:          .word 0
D:          .word 0
C:          .word 0
B:          .word 0
A:          .word 0
var1:       .word 0
var2:       .word 0
var3:       .word 0
quotient:   .word 0
remainder:  .word 0

        .text
        .globl main
main:
# ========== Step 1 ==========
        # Step 1.1 # Display a prompt, "Please enter the value E: "
        li      $v0, 4
        la      $a0, promptE
        syscall
        # Step 1.2 # Setup to wait for the user to enter a non-zero positive integer.
        li      $v0, 5
        syscall
        # Step 1.3 # Store the entered value to the variable E defined in the data segment.
        sw      $v0, E

# ========== Step 2 ==========
        # Step 2.1 # Display a prompt, "Please enter the value D: "
        li      $v0, 4
        la      $a0, promptD
        syscall
        # Step 2.2 # Setup to wait for the user to enter a non-zero positive integer
        li      $v0, 5
        syscall
        # Step 2.3 # Store the entered value to the variable D defined in the data segment.
        sw      $v0, D

# ========== Step 3 ==========
        # Step 3.1 # Display a prompt, "Please enter the value C: "
        li      $v0, 4
        la      $a0, promptC
        syscall
        # Step 3.2 # Setup to wait for the user to enter a non-zero positive integer
        li      $v0, 5
        syscall
        # Step 3.3 # Store the entered value to the variable C defined in the data segment.
        sw      $v0, C

# ========== Step 4 ==========
        # Step 4.1 # Display a prompt, "Please enter the value B: "
        li      $v0, 4
        la      $a0, promptB
        syscall
        # Step 4.2 # Setup to wait for the user to enter a non-zero positive integer
        li      $v0, 5
        syscall
        # Step 4.3 # Store the entered value to the variable B defined in the data segment.
        sw      $v0, B

# ========== Step 5 ==========
        # Step 5.1 # Display a prompt, "Please enter the value A: "
        li      $v0, 4
        la      $a0, promptA
        syscall
        # Step 5.2 # Setup to wait for the user to enter a non-zero positive integer
        li      $v0, 5
        syscall
        # Step 5.3 # Store the entered value to the variable A defined in the data segment.
        sw      $v0, A

# ========== Step 6 ==========
        # Step 6.1 # Load the value stored in variable B and C to two registers.
        lw      $t0, B
        lw      $t1, C
        # Step 6.2 # Add the values in the two registers.
        add     $t2, $t0, $t1
        # Step 6.3 # Store the addition result to the variable var1.
        sw      $t2, var1

# ========== Step 7 ==========
        # Step 7.1 # Load the value stored in variable A to a register.
        lw      $t3, A
        # Step 7.2 # Multiply the value in var1 to the register.
        mul     $t4, $t2, $t3
        # Step 7.3 # Store the lower 32-bit of the multiplication result to the variable var2.
        sw      $t4, var2

# ========== Step 8 ==========
        # Step 8.1 # Load the values stored in variable D to a register.
        lw      $t5, D
        # Step 8.2 # Left shift the register value (by 4).
        sll     $t5, $t5, 4
        # Step 8.3 # Subtract the bit shifted result from var2.
        sub     $t6, $t4, $t5
        # Step 8.4 # Store the subtraction result to the variable var3.
        sw      $t6, var3

# ========== Step 9 ==========
        # Step 9.1 # Load the value stored in variable E to a register.
        lw      $t7, E
        # Step 9.2 # Divide the var3 in the register by the variable E.
        div     $t6, $t7  # lo: quotient, hi: remainder
        # Step 9.3 # Store the quotient of the division result to the variable quotient.
        mflo    $t8
        sw      $t8, quotient
        # Step 9.4 # Store the remainder of the division result to the variable remainder.
        mfhi    $t9
        sw      $t9, remainder

        # Step 9.5 # Print all the intermediate results according to the sample output format.
        li      $v0, 4
        la      $a0, msg_var1
        syscall
        li      $v0, 1
        lw      $a0, var1
        syscall
        li      $v0, 4
        la      $a0, newline
        syscall

        li      $v0, 4
        la      $a0, msg_var2
        syscall
        li      $v0, 1
        lw      $a0, var2
        syscall
        li      $v0, 4
        la      $a0, newline
        syscall

        li      $v0, 4
        la      $a0, msg_var3
        syscall
        li      $v0, 1
        lw      $a0, var3
        syscall
        li      $v0, 4
        la      $a0, newline
        syscall

        li      $v0, 4
        la      $a0, msg_quot
        syscall
        li      $v0, 1
        lw      $a0, quotient
        syscall
        li      $v0, 4
        la      $a0, newline
        syscall

        li      $v0, 4
        la      $a0, msg_rem
        syscall
        li      $v0, 1
        lw      $a0, remainder
        syscall
        li      $v0, 4
        la      $a0, newline
        syscall

        # Step 9.6 # Print the Author message.
        li      $v0, 4
        la      $a0, author
        syscall

        # Step 9.7 # Exit from the program.
        li      $v0, 10
        syscall

Important: Replace [Your Name, Class, Student ID] in the author string with your own details before submission.

4. Detailed Explanation of Each Solution Step

4.1 Data Section (`.data`)

Prompts (promptE, promptD, …): .asciiz strings for “Please enter the value E: ”, etc. Syscall 4 prints the string whose address is in $a0.
Result messages (msg_var1, …): Strings like "var1 = (B + C) = " so the output matches the required format.
Variables (E, D, C, B, A, var1, var2, var3, quotient, remainder): .word 0 reserves one word (4 bytes) per variable. Inputs are stored here after each read; computed values are stored here as specified in the pseudo-code.

4.2 Steps 1–5: Reading E, D, C, B, A

Each step does the same pattern:

Print prompt: li $v0, 4, la $a0, promptX, syscall — MARS syscall 4 = “print string”; the string address must be in $a0.
Read integer: li $v0, 5, syscall — syscall 5 = “read integer”; the value is returned in $v0.
Store to memory: sw $v0, E (or D, C, B, A) — store the word in $v0 at the label E (or D, C, B, A).

We read in order E, D, C, B, A so that the user supplies values from smallest to largest, matching the constraint E < D < C < B < A.

4.3 Step 6: var1 = B + C

6.1 Load B and C: lw $t0, B and lw $t1, C. Now $t0 = B, $t1 = C.
6.2 Add: add $t2, $t0, $t1 → $t2 = B + C.
6.3 Store: sw $t2, var1 → var1 = B + C.

add is R-type; result is in the destination register. We keep $t2 for the next step (var1 is also needed for var2).

4.4 Step 7: var2 = var1 × A (lower 32 bits)

7.1 Load A: lw $t3, A → $t3 = A. (We already have var1 in $t2.)
7.2 Multiply: mul $t4, $t2, $t3 → $t4 = var1 × A. In MIPS32, mul puts the low 32 bits of the product in the destination register, which satisfies “only consider the lower 32-bit of the multiplication result”.
7.3 Store: sw $t4, var2 → var2 = var1 * A.

4.5 Step 8: var3 = var2 − (D ≪ 4)

8.1 Load D: lw $t5, D → $t5 = D.
8.2 Left shift by 4: sll $t5, $t5, 4 → $t5 = D × 16 (D ≪ 4). sll is R-type; the shift amount is 4; result overwrites the register.
8.3 Subtract: sub $t6, $t4, $t5 → $t6 = var2 − (D ≪ 4). We use $t4 (var2) and the shifted value in $t5.
8.4 Store: sw $t6, var3 → var3 = var2 - (D << 4).

4.6 Step 9.1–9.4: Quotient and Remainder

9.1 Load E: lw $t7, E → $t7 = E.
9.2 Divide: div $t6, $t7 — divides $t6 (var3) by $t7 (E). MIPS div puts quotient in lo and remainder in hi.
9.3 Quotient: mflo $t8 copies lo into $t8; then sw $t8, quotient stores it in quotient.
9.4 Remainder: mfhi $t9 copies hi into $t9; then sw $t9, remainder stores it in remainder.

So we get quotient = var3 / E and remainder = var3 % E using only course instructions.

4.7 Step 9.5: Printing Intermediate Results

For each of var1, var2, var3, quotient, and remainder we:

Print a label string: syscall 4 with the address of the corresponding message.
Print the number: syscall 1 with the value in $a0 (load from the variable with lw $a0, var1 etc.).
Print a newline: syscall 4 with newline.

This produces output in the form:

var1 = (B + C) = 15
var2 = var1 * A = 135
etc.

4.8 Step 9.6–9.7: Author Message and Exit

9.6 Print the author line: syscall 4 with la $a0, author. Remember to set the author string in .data to your name, class, and student ID.
9.7 Exit: li $v0, 10, syscall — MARS syscall 10 terminates the program. The lab requires the program termination message to be shown in the Run I/O window.

5. Summary Table: Instructions Used

Task	MIPS instructions / syscalls used
Print string	`li $v0, 4`, `la $a0, label`, `syscall`
Read integer	`li $v0, 5`, `syscall`; result in `$v0`
Store input	`sw $v0, E` (or D, C, B, A)
Load from memory	`lw reg, label`
Add	`add rd, rs, rt`
Multiply (low 32)	`mul rd, rs, rt`
Subtract	`sub rd, rs, rt`
Left shift by 4	`sll rd, rt, 4`
Division	`div rs, rt` → quotient in `lo`, remainder in `hi`
Get quotient	`mflo rd`
Get remainder	`mfhi rd`
Print integer	`li $v0, 1`, `lw $a0, var`, `syscall`
Exit	`li $v0, 10`, `syscall`

6. Checklist for Submission

[ ] Fill in name, class, student ID (and photo if required) on the answer sheet.
[ ] Use the exact prompts and formula from the lab (E, D, C, B, A order; var1, var2, var3, quotient, remainder).
[ ] Replace the author string with your details.
[ ] Run and single-step the program in MARS to verify (e.g. E=5, D=6, C=7, B=8, A=9 → var1=15, var2=135, var3=39, quotient=7, remainder=4).
[ ] Include in the report: source code listing, assembled machine code (Text Segment), and Run I/O output including program termination.
[ ] Submit both the lab report (PDF) and the .asm file with the required naming.

Using this solution and explanations, you can complete Lab 03 and understand each step in detail.

ICO Lab 03 — Answers and Detailed Solutions

1. Understanding the Problem

1.1 Formula

1.2 Constraint on Inputs

1.3 Variables to Compute and Store (in data memory)

2. Manual Calculation (Sample)

3. MIPS Solution — Complete Assembly Program

4. Detailed Explanation of Each Solution Step

4.1 Data Section (`.data`)

4.2 Steps 1–5: Reading E, D, C, B, A

4.3 Step 6: var1 = B + C

4.4 Step 7: var2 = var1 × A (lower 32 bits)

4.5 Step 8: var3 = var2 − (D ≪ 4)

4.6 Step 9.1–9.4: Quotient and Remainder

4.7 Step 9.5: Printing Intermediate Results

4.8 Step 9.6–9.7: Author Message and Exit

5. Summary Table: Instructions Used

6. Checklist for Submission

Artículos relacionados

Mobile Application Development CCIT4059

Mobile Application Development CCIT4059

COMSM1302 電腦架構概論四份模擬試卷完整詳解

COMSM1302 電腦架構概論卡諾圖化簡 15 題精練

COMSM1302 電腦架構概論 Sheet 1 完整詳解

COMSM1302 電腦架構概論第一週完整教材：數位邏輯基礎

1. Understanding the Problem

1.1 Formula

1.2 Constraint on Inputs

1.3 Variables to Compute and Store (in data memory)

2. Manual Calculation (Sample)

3. MIPS Solution — Complete Assembly Program

4. Detailed Explanation of Each Solution Step

4.1 Data Section (.data)

4.2 Steps 1–5: Reading E, D, C, B, A

4.3 Step 6: var1 = B + C

4.4 Step 7: var2 = var1 × A (lower 32 bits)

4.5 Step 8: var3 = var2 − (D ≪ 4)

4.6 Step 9.1–9.4: Quotient and Remainder

4.7 Step 9.5: Printing Intermediate Results

4.8 Step 9.6–9.7: Author Message and Exit

5. Summary Table: Instructions Used

6. Checklist for Submission

Artículos relacionados

Mobile Application Development CCIT4059

Mobile Application Development CCIT4059

COMSM1302 電腦架構概論四份模擬試卷完整詳解

COMSM1302 電腦架構概論卡諾圖化簡 15 題精練

COMSM1302 電腦架構概論 Sheet 1 完整詳解

COMSM1302 電腦架構概論第一週完整教材：數位邏輯基礎

4.1 Data Section (`.data`)