THE UNIVERSITY OF HONG KONG Department of Computer Science COMP2120B Computer Organization Assignment 2
Due Date: Sun Mar 14, 2021.
Consider a simple 32-bit processor with the data path as shown in fig. 1. The processor has 32 general purpose registers. There are 3 buses, S1-bus, S2-bus and D-bus connecting the registers for data movement. The register files has 2 read ports and 1 write port (i.e. it can perform 2 read and 1 write at the same time).
D-Bus
”
uEE
S1-Bus S2-Bus
”u ”u
EE
EE
C
ALU
B
Register File
T
PC
uEE E
RFOUT1
‘
EuE
EE
IR
T
‘u c
Figure 1: A simplified CPU
The processor has instructions which specifies 3 operands explicitly (namely, 2 source and 1 destination operands). The leftmost byte of the instruction represents the oper- ation to be performed, such as ADD, SUB etc. For arithmetic and logic operations, the
1
‘
A
u
u
RFIN
RFOUT2
IncPC
MAR
MBR
External Memory
operands must be in registers. Hence the 3 bytes will give the addresses of operands in the register file. There will be a direct path connecting these 3 bytes in the IR (Instruc- tion Register) to the address of the register file, so that when you perform read/write on register file, the register specified in these bytes will be accessed.
If the instruction is LOAD or STORE to load a word from memory to register, and vice versa, the source operand (LOAD) or destination operand (STORE) refer to a memory address. How to find this address is specified by Addressing Mode. In this machine, for simiplicity, the memory operand byte (source/destination) will always be 1111 1111 (or in hex 0xff), which means that the actual 32-bit memory address will be given in the word following the instruction (see example program below).
The ALU has the following operations: ADD, SUB, bitwise AND, OR, and NOT. For operations with only one operand (e.g. NOT), source operand 1 is used, and source operand 2 is empty.
Finally, there is a branch instruction, which performs conditional or unconditional branch as specified in the cc field of the instruction. The branch address is specified in the word following the instruction, the same as in LOAD/STORE instruction.
Instruction Format Arithmetic/Logic Instruction
The instruction format of the machine (except LOAD, STORE and BRANCH):
The instructions can be categorized into the following types:
Arithmetic Operations
ADD R1, R2, R3 ; R3 <- R1 + R2
SUB R1, R2, R3 ; R3 <- R1 - R2
Opcode
Source Operand 1
Source Operand 2
Destination Operand
Logical Operations
AND R1, R2, R3
OR R1, R2, R3
NOT R1, R3
;R3<-R1andR2 ;R3<-R1orR2 ;R3<-notR1
Data Movement Instruction
MOV R1, R3 ; R3 <- R1
Note that in the NOT and MOV operation, source operand 2 field is not used and will be set as 00000000.
2
Load/Store Instruction
Moving data from Memory to registers and vice versa.
LD A, R3 ; R3 <- A, A is in memory
ST R1, A ; A <- R1, A is in memory
Load instruction:
Store instruction:
where the addressing mode (how to find the target address) is specified in byte 2 of the instruction. In this machine, only one addressing mode is used, where the target address is given by the word following the LOAD or STORE instruction (Absolute Addressing). This is specified as 11111111 in that byte.
Control Instruction
Control flow is by using BRANCH instruction. There are two types of branch instruction — conditional and unconditional Branch. Branch Instruction Format:
Conditional branch is based on the result of previous ALU operation, which is store in a flag register. In this machine, we only use a ZERO flag, which will be set to 1 if the ALU operation results in 0, and set to 0 otherwise. The target address is specified in the same way as in memory operation. Similarly, the byte of Addressing Mode is set to 11111111.
The condition code is specified as
Halt Instruction
The HLT instruction is used to stop the program. The other 3 bytes are all 0.
Opcode (Load)
00000000
Addressing Mode
Destination Operand
Opcode (Store)
Source Operand
Addressing Mode
00000000
Opcode (Branch)
Condition Code (cc)
Addressing Mode
00000000
Condiation Code (cc)
Instruction
Description
00000000
BR
Unconditaion Branch, always goto
00000001
BZ
Branch if Zero flag is set
00000010
BNZ
Branch if Zero flag is NOT set
3
Opcodes
Instruction
Opcode
Instruction
Opcode
Instruction
Opcode
ADD
00000000
OR
00000100
Bcc
00001000
SUB
00000001
MOV
00000101
HLT
00001001
NOT
00000010
LD
00000110
AND
00000011
ST
00000111
Part I: Example Program
The simulator program is given in sim.py. The code for the SUB and ST instruction is missing. 问题1:Study the simulator code carefully, and complete the missing part.
Running the simulator program:
[python3] sim.py [-d] prog
If -d option is specified, the program will print out debug information. The simulator obtains input program from the file prog.
Test you simulator with the following simple program:
LD P0,R4
LD P1,R1
MOV R1,R2
LD P2,R3
0000: 0600ff04
0008: 0600ff01
0010: 05010002
0014: 0600ff03
001C: 00040104
0020: 00010201
0024: 01030105
0028: 0802ff00
0030: 0704ff00
0038: 09000000
003C: 00000000
0040: 00000001
0044: 0000000a
0048: 00000000
0000003c
00000040
00000044
0000001c
00000048
L: ADD
ADD R1,R2,R1
SUB R3,R1,R5
BNZ L
ST R4,P
HLT
P0: .WORD
P1: .WORD
P2: .WORD
P: .WORD
0 1 A
R4,R1,R4
What does this program do?高斯求和 Part II: Hand Assemble
问题2:Translate the following program into hexadecimal form, and put it in a file named prog2 with the same format as the file prog. Run the simulator by
[python3] sim.py [-d] prog2
Write down the final result stored in P.P = 20 What does the program do?乘法
4
LD P0, R4
LD P1, R1
LD P2, R2
LD P3, R3
L: ADD
SUB R3, R1, R3
BNZ L
ST R4, P
HLT
P0: .WORD 0
P1: .WORD 1
P2: .WORD 5
P3: .WORD 4
P: .WORD
A working simulator (executable pyc file only, without source code) is given to you, so that you can complete this part using this program, if your simulator in Part I is not working.
R4, R2, R4
5