Assignment 1 Answers — Introduction to Computer Organization CCIT 4026

Student Number: 20285660
HKU SPACE Community College — AY2526 S2

Digit Extraction from Student Number 20285660

From 20285660:

Position	Digit	Usage
1st (rightmost)	0	Starting address of data segment (Q2) → 0x000000A0
2nd (second-rightmost)	6	Instruction pair selection (Q3)
Last 5 digits (d4 d3 d2 d1 d0)	8, 5, 6, 6, 0	Q2 values
5th–8th digits (a0, a1, b0, b1)	5, 6, 6, 0	Q4 array values

Q1. [30 marks] Memory Map of Data Segment

Important Note

The assignment PDF states “Given the following data segment definition” but the specific .data segment code was not visible in the extracted text (it may appear in an image or formatted block). Use the exact data segment provided in your assignment handout and apply the method below.

Method: Building the Memory Map

Given: - Base address: 0x2000 - Endianness: Little Endian (least significant byte at lowest address) - Initial state: All memory = 0 before data is loaded - Convention: Mark reserved (unused) bytes with X

MIPS Data Directives and Sizes

Directive	Size	Example
`.space n`	n bytes	`.space 2` → 2 bytes
`.byte`	1 byte	`.byte 0x12` → 1 byte
`.half`	2 bytes	`.half 0x3456` → 2 bytes
`.word`	4 bytes	`.word 0x12345678` → 4 bytes
`.asciiz "str"`	strlen + 1 bytes	`"Hi"` → 3 bytes

Little Endian Layout

For multi-byte values, the least significant byte is stored at the lowest address:

Example: 0x12345678
0x2000: 0x78 (LSB)
0x2001: 0x56
0x2002: 0x34
0x2003: 0x12 (MSB)

Given Data Segment (from assignment)

.data
.space 3
Str1: .ascii "abc"
.align 2
Str2: .asciiz "DEFG"
Hf:   .half 18
W:    .word -1, 3
B:    .byte 4
Str3: .asciiz "\n"

Layout:

Byte Address	Label	Content	Explanation
0x2000	X	X	`.space 3` first byte
0x2001	X	X	`.space 3` second byte
0x2002	X	X	`.space 3` third byte
0x2003	Str1	0x61	'a'
0x2004	Str1	0x62	'b'
0x2005	Str1	0x63	'c'
0x2006	Str2	0x44	'D'
0x2007	Str2	0x45	'E'
0x2008	Str2	0x46	'F'
0x2009	Str2	0x47	'G'
0x200A	Str2	0x00	null
0x200B	Hf	0x12	18 LSB
0x200C	Hf	0x00	18 MSB
0x200D	W	0xFF	-1 byte 0
0x200E	W	0xFF	-1 byte 1
0x200F	W	0xFF	-1 byte 2
0x2010	W	0xFF	-1 byte 3
0x2011	W	0x03	3 byte 0
0x2012	W	0x00	3 byte 1
0x2013	W	0x00	3 byte 2
0x2014	W	0x00	3 byte 3
0x2015	B	0x04	4
0x2016	Str3	0x0A	newline
0x2017	Str3	0x00	null
0x2018	—	0x00	(beyond data)
0x2019	—	0x00	(beyond data)
0x201A	—	0x00	(beyond data)
0x201B	—	0x00	(beyond data)

Place X for reserved bytes (0x2000–0x2002). Addresses 0x2018–0x201B are beyond the defined data; content remains 0 per the initialization assumption.

Q2. [20 marks] Register Values After Program Execution

Student-Specific Values (20285660)

1st student number: 0 → data segment base = 0x000000A0
d4 d3 d2 d1 d0 = 8, 5, 6, 6, 0

Data Segment Layout

.data
sid: .word d2 d1 d0    # Three consecutive words
.word d4               # One more word
s:   .word d3          # One word

Address	Content	Label
0xA0 (sid+0)	d2 = 6	sid
0xA4 (sid+4)	d1 = 6
0xA8 (sid+8)	d0 = 0
0xAC (sid+12)	d4 = 8
0xB0 (s)	d3 = 5	s

Step-by-Step Trace

Instruction	Operation	Register Values
`la $t0, sid`	Load address of `sid`	$t0 = 0xA0
`lw $t1, 8($t0)`	Load word at sid+8	$t1 = mem[0xA8] = d0 = 0
`la $t2, s`	Load address of `s`	$t2 = 0xB0 = 160
`lw $t3, -4($t2)`	Load word at s−4 = 0xAC	$t3 = mem[0xAC] = d4 = 8
`li $t4, 0x20`	Load immediate 0x20	$t4 = 32
`add $t4, $t4, $t1`	$t4 = 32 + 0	$t4 = 32
`sub $t5, $t4, $t3`	$t5 = 32 − 8	$t5 = 24

Final Register Values (Decimal)

Register	Value	Derivation
$t1	0	d0 from sid+8
$t2	160	Address of s (0xB0)
$t3	8	d4 from s−4
$t4	32	0x20 + $t1 = 32 + 0
$t5	24	$t4 − $t3 = 32 − 8

Q3. [25 marks] Machine Code Encoding

Instruction Selection (2nd student number = 6)

From the assignment table, for 2nd student number = 6:

Instruction 1: add $s5, $t5, $v1
Instruction 2: lw $t1, -12($s1)

Instruction 1: `add $s5, $t5, $v1`

Format: R-type

Fields:

Field	Value	Registers / Meaning
op	000000	R-type
rs	$t5 (21)	First source
rt	$v1 (3)	Second source
rd	$s5 (21)	Destination
shamt	00000	Not used
funct	100000	`add`

Binary:

000000  10101  00011  10101  00000  100000
op      rs     rt     rd     shamt  funct

HEX: 0x02B5A820

Check: op=0, rs=21, rt=3, rd=21, funct=32.

Instruction 2: `lw $t1, -12($s1)`

Format: I-type
Layout: op | rs | rt | immediate

op: 100011 (35) — load word
rs: $s1 (17) — base register
rt: $t1 (9) — destination
immediate: −12 = 0xFFF4 (16-bit two’s complement)

Two’s complement for −12: - 12 in binary: 0000 0000 0000 1100
- Invert: 1111 1111 1111 0011
- Add 1: 1111 1111 1111 0100 = 0xFFF4

Binary:

100011  10001  01001  1111111111110100
op      rs     rt    immediate

HEX: 0x8E29FFF4

Check: op=0x23, rs=17, rt=9, imm=0xFFF4.

Q3 Summary

Instruction	Machine Code (HEX)
`add $s5, $t5, $v1`	0x02B5A820
`lw $t1, -12($s1)`	0x8E29FFF4

Q4. [25 marks] Dot Product Program

Values from Student Number 20285660

For student number 20285660, the 5th–8th digits are: 5, 6, 6, 0

a0 = 5, a1 = 6, b0 = 6, b1 = 0

Dot Product Calculation

[ C = \langle A \cdot B \rangle = \sum_{i=0}^{1} a_i b_i = a_0 b_0 + a_1 b_1 ]

[ C = 5 * 6 + 6 * 0 = 30 + 0 = 30 ]

Expected Output

The dot product C = 30

MIPS Program

# Dot Product Program
# Student Number: 20285660
# a0=5, a1=6, b0=6, b1=0
# C = 5*6 + 6*0 = 30

.data
Array_A: .word 5, 6      # a0, a1
Array_B: .word 6, 0      # b0, b1
msg:     .asciiz "The dot product C = "
newline: .asciiz "\n"

.text
.globl main
main:
    # Load array base addresses
    la $t0, Array_A
    la $t1, Array_B

    # Compute a0 * b0
    lw $t2, 0($t0)       # a0
    lw $t3, 0($t1)       # b0
    mul $t4, $t2, $t3    # a0 * b0

    # Compute a1 * b1
    lw $t2, 4($t0)       # a1
    lw $t3, 4($t1)       # b1
    mul $t5, $t2, $t3    # a1 * b1

    # C = a0*b0 + a1*b1
    add $t6, $t4, $t5    # $t6 = C

    # Print "The dot product C = "
    li $v0, 4
    la $a0, msg
    syscall

    # Print C
    li $v0, 1
    move $a0, $t6
    syscall

    # Print newline
    li $v0, 4
    la $a0, newline
    syscall

    # Exit
    li $v0, 10
    syscall

Program Logic

Addresses: $t0 → Array_A, $t1 → Array_B.
a0×b0: Load A[0], B[0], multiply, store in $t4.
a1×b1: Load A[1], B[1], multiply, store in $t5.
C: Add $t4 and $t5 into $t6.
Print: Syscall 4 for the message, syscall 1 for C, then newline.
Exit: Syscall 10.

Note: Run this in MARS and capture the source code and “Run I/O” output for your submission.

Quick Reference Summary

Question	Key Values	Answer
Q1	Base 0x2000, Little Endian	See complete memory map table above (Str1, Str2, Hf, W, B, Str3)
Q2	d4=8, d3=5, d2=6, d1=6, d0=0; base 0xA0	$t1=0, $t2=160, $t3=8, $t4=32, $t5=24
Q3	2nd digit = 6	Instruction 1: `0x02B5A820`; Instruction 2: `0x8E29FFF4`
Q4	a0=5, a1=6, b0=6, b1=0	C = 30

Assignment 1 Answers — Introduction to Computer Organization CCIT 4026

Digit Extraction from Student Number 20285660

Q1. [30 marks] Memory Map of Data Segment

Important Note

Method: Building the Memory Map

MIPS Data Directives and Sizes

Little Endian Layout

Given Data Segment (from assignment)

Q2. [20 marks] Register Values After Program Execution

Student-Specific Values (20285660)

Data Segment Layout

Step-by-Step Trace

Final Register Values (Decimal)

Q3. [25 marks] Machine Code Encoding

Instruction Selection (2nd student number = 6)

Instruction 1: `add $s5, $t5, $v1`

Instruction 2: `lw $t1, -12($s1)`

Q3 Summary

Q4. [25 marks] Dot Product Program

Values from Student Number 20285660

Dot Product Calculation

Expected Output

MIPS Program

Program Logic

Quick Reference Summary

Related articles

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

Related articles

Computer Organization Updated 2026-05-13

Mobile Application Development CCIT4059

使用方法：依時間表逐節閱讀。每節有「核心概念 → 重點整理 → 易考點 → 自我檢測」。建議準備白紙邊看邊抄畫。
Read article →

Computer Organization Updated 2026-05-13

Mobile Application Development CCIT4059

涵蓋 Unit 1–10 全部內容。每題含答案、詳細解答過程、對應單元、易錯提示。建議：先自己作答，再對照解析。
Read article →

Computer Organization Updated 2026-07-10

COMSM1302 電腦架構概論四份模擬試卷完整詳解

, frame=single, rulecolor=, breaklines=true, columns=fullflexible, keepspaces=true, xleftmargin=6pt, xrightmargin=6pt, commentstyle=, morecomment=[l]//,
Read article →

Computer Organization Updated 2026-07-14

COMSM1302 電腦架構概論卡諾圖化簡 15 題精練

HTML1E7D32 HTMLEDF7ED HTML1A56A0 HTMLB71C1C HTML5E35B1 HTMLC8E6C9 HTMLFFE0B2 HTMLBBDEFB
Read article →

Computer Organization Updated 2026-07-10

COMSM1302 電腦架構概論 Sheet 1 完整詳解

HTML1E7D32 HTMLEDF7ED HTML1A56A0 HTMLF5F5F5 HTMLB71C1C HTML5E35B1 HTMLC8E6C9 HTMLFFE0B2 HTMLBBDEFB
Read article →

Computer Organization Updated 2026-07-10

COMSM1302 電腦架構概論第一週完整教材：數位邏輯基礎

HTML1E7D32 HTMLEDF7ED HTML1A56A0 HTMLF5F5F5 HTMLB71C1C HTML5E35B1 HTMLBBDEFB
Read article →

Digit Extraction from Student Number 20285660

Q1. [30 marks] Memory Map of Data Segment

Important Note

Method: Building the Memory Map

MIPS Data Directives and Sizes

Little Endian Layout

Given Data Segment (from assignment)

Q2. [20 marks] Register Values After Program Execution

Student-Specific Values (20285660)

Data Segment Layout

Step-by-Step Trace

Final Register Values (Decimal)

Q3. [25 marks] Machine Code Encoding

Instruction Selection (2nd student number = 6)

Instruction 1: add $s5, $t5, $v1

Instruction 2: lw $t1, -12($s1)

Q3 Summary

Q4. [25 marks] Dot Product Program

Values from Student Number 20285660

Dot Product Calculation

Expected Output

MIPS Program

Program Logic

Quick Reference Summary

Related articles

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

Instruction 1: `add $s5, $t5, $v1`

Instruction 2: `lw $t1, -12($s1)`