1 Introduction

As discussed in class, register-transfer level design can be relatively simply described as

1. clock signal has a rising edge
2. register outputs change
3. logic works, which can be thought of as arbitrary acyclic code that assigns each variable only once
4. logic results in register inputs changing
5. repeat

In this lab, and the subsequent programming assignment, you will expand on these basic pieces to build your own machine simulator and write binary code for it.

2 Getting the simulator skeleton

We’ve written a basic simulator in python (sim_base.py) and java (SimBase.java). This works from a command-line interface, expecting memory byte values (either directly or in a file) as a command-line argument:

# runs SimBase in java with 8 bytes of memory set
            javac SimBase.java
            java SimBase 01 23 45 67 89 ab cd ef

# runs SimBase in python with 8 bytes of memory set
            python3 sim_base.py 01 23 45 67 89 ab cd ef

# runs SimBase using the contents of memory.txt to set memory
            python3 sim_base.py memory.txt
            java SimBase memory.txt

Memory contents must be specified in hexadecimal bytes, separated by whitespace.

3 Write the simulator functionality

Each file begins with a function or method named execute which is given two arguments (the current instruction in ir and the PC of this instruction in oldPC) and returns one value (the pc to execute next). The skeleton code just returns oldPC + 1.

Each file also has two global values you can access: R, an array of 4 register values, and M, an array of 256 memory values.

We also provide a helper function for getting a range of bits from a number.

3.1 Language nuances

Python’s syntax for !x is not x instead.

Java treats bytes (like R[i]) as signed integers, not unsigned. That means they are not good indices (e.g., M[R[i]] might throw an exception if R[i] is negative). However, R[i] & 0xFF treats it as unsigned instead, so M[R[i] & 0xFF] should work.

Python treats bytes as unsigned, so R[i] <= 0 is always true. It can be re-written to work correctly as R[i] == 0 or R[i] >= 0x80

3.2 Example programs

Halt

The code

00 00 00 80

should advance to pc 3 and then stay there.

Try finding other code that will also do that.

Move

Consider the following code

Bytes	Activity
68 23	move 0x23 into register 2
06	move from register 2 to register 1
60 20	move 0x20 into register 0
44	move from register 1 to memory at address from register 0 (i.e., move 0x23 to address 0x20)
80	halt

If your simulator works, then 68 23 06 60 20 44 80 should end with registers being [20, 23, 23, 00] and a 0x23 in memory at address 0x20.

Try adding code that will read from memory into register 3.

Math

Consider the following code

Bytes	Activity
68 10	move 0x10 into R2
32	move value from address R2 into R0
68 11	move 0x11 into R2
36	move value from address R2 into R1
14	R1 += R0
68 12	move 0x12 into R2
46	move R1 into address R2
80	halt
five 00s	(padding)
23 A1	some numbers to add

If your simulator works, you should end up with 0xC4 (i.e., 0x23 + 0xA1) in address 0x12.

Try adding code that can sum more numbers from memory, not just those two.

Conditional jump

The following should not jump, so three steps should end up with the PC at address 5, not 20:

pseudocode	parts	bytes
R0 = 10	(6, 0, 0) 10	60 0A
R1 = 20	(6, 1, 0) 20	64 14
if R0 <= 0, jump to R1	(7, 0, 1)	71

The following should jump, so three steps should end with the PC at address 20, not 5:

pseudocode	parts	bytes
R0 = −10	(6, 0, 0) −10	60 F6
R1 = 20	(6, 1, 0) 20	64 14
if R0 <= 0, jump to R1	(7, 0, 1)	71

Read from immediate address

The following should load 20 into R₁

pseudocode	parts	bytes
R2 = 10	(6, 2, 0) 10	68 0A
R3 = 20	(6, 3, 0) 20	6C 14
write R3 to address R2	(4, 3, 2)	4E
read address 10 into R1	(6, 1, 3) 10	67 0A

You could also do this without using instruction 4 by setting enough memory that there was already data in address 10:

pseudocode	parts	bytes
read address 10 into R1	(6, 1, 3) 10	67 0A
intialize address 10 with 20	0s, then 20	00 00 00 00 00 00 00 14

3.3 More, less-explained examples

icode = 0

First load a value into R₁ then set R₂ to equal R₁: 64 14 09

icode = 1

First load a value into R₁ and R₂ then add them together: 64 14 68 20 19

icode = 2

First load a value into R₁ and R₂ then and them together: 64 14 68 20 29

icode = 3

First load a value into R₁ and then load that memory address into R₂: 64 02 39

icode = 4

First load a value into R₁ and R₂ and then store R₁ into memory at R₂: 64 02 68 20 46

icode = 5

First load a value into R₁ and then do something to it:

• flip all bits: 64 89 54 (result: 76)
• perform !: 64 89 55 (result: 00; try also 55 by itself to result in 01)
• negate: 64 89 56 (result: 77)
• replace with the PC: 64 89 57 (result: 02)

icode = 6

First load a value into R₁ and then use an immediate:

• replace with immediate: 64 14 64 20 (result: 20)
• add immediate: 64 14 65 20 (result: 34)
• and immediate: 64 14 66 24 (result: 04)
• replace with memory contents at immediate: 64 14 67 02 (result: 67)

icode = 7

First load an address into R₁, a value into R₂, and then conditionally jump:

• jumps: 64 14 68 ff 79
• jumps: 64 14 68 80 79
• jumps: 64 14 68 00 79
• does not jump: 64 14 68 01 79
• does not jump: 64 14 68 7f 79

4 Check-off

To check-off this lab, show a TA your simulator and the memory contents that do the tasks listed above.