Description

Prof. Ramachandran & Prof. Daglis
Homework 2 – Calling Convention
1 Problem 1: Getting Started with the LC-5500
In this homework, you will be using the LC-5500 ISA to complete a Tower of Hanoi move-counting function. Before you begin, you should familiarize yourself with the available instructions, the register conventions and the calling convention of LC-5500. Details can be found in the section, Appendix A: LC-5500 Instruction Set Architecture, at the end of this document.
The assembly folder contains several tools for you to use:
• assembler.py: a basic assembler that can take your assembly code and convert it into binary instructions for the LC-5500.
• lc5500.py: the ISA definition file for the assembler, which tells assembler.py the instructions supported by the LC-5500 and their formats.
• lc5500-sim.py: A simulator of the LC-5500 machine. The simulator reads binary instructions and emulates the LC-5500 machine, letting you single-step through instructions and check their results.
To learn how to run these tools, see the README.md file in the assembly directory.
Before you begin work on the second problem of the homework, try writing a simple program for the LC-5500 architecture. This should help you familiarize yourself with the available instructions.
We have provided a template, mod.s, for you to use for this purpose. Try writing a program that performs the mod operation on the two provided arguments. A correct implementation will result in a value of 2.
You can use the following C code snippet as a guide to implement this function:
int mod(int a, int b) {
int x = a; while (x >= b) {
x = x – b;
}
return x;
}
There is no turn-in for this portion of the assignment, but it is recommended that you attempt it in order to familiarize yourself with the ISA.
2 Problem 2: Tower of Hanoi
For this problem, you will be implementing the missing portions of the program that calculates the minimum number of moves to solve the Tower of Hanoi problem for n disks.
You will be finishing a recursive implementation of the Tower of Hanoi minimal moves calculator program that follows the LC-5500 calling convention. Recursive functions always obtain a return address through the function call and return to the callee using the return address.
You must use the stack pointer ($sp) and frame pointer ($fp) registers as described in the textbook and lecture slides. Use the PUSH and POP instructions to manipulate the stack.
Here is the C code for the Tower of Hanoi minimal moves calculator you have been provided:
int minimumHanoi(int n) {
if (n == 1) return 1;
else
return (2 * minimumHanoi(n – 1)) + 1;
}
Note that this C code is just to help your understanding and does not need to be exactly followed. However, your assembly code implementation should meet all of the given conditions in the description.
Open hanoi.s file in the assembly directory. This file contains an implementation of the Tower of Hanoi minimal moves calculator program that is missing significant portions of the calling convention. Near the bottom of the hanoi.s we have provided multiple numbers that you can use to test your homework. They are located at labels testNumDisks1, testNumDisks2, testNumDisks3. Be sure to use these provided integers by loading them from the labels into registers. None of the numbers provided and tested will be lower than
1.
Complete the program by implementing the various missing portions of the LC-5500 calling convention. Each location where you need to implement a portion of the calling convention is marked with a TODO label as well as a short hint describing the portion of the calling convention you should be implementing.
Please note that we will be testing your implementation for multiple different instances, so please do not attempt to hardcode your solutions.
3 Problem 3: Short Answer
Please answer the following question in the file named answers.txt:
4 Deliverables
• hanoi.s: your assembly code from Section 2
• answers.txt: your answer to the problem from Section 3
The TAs should be able to type python assembler.py -i lc5500 –sym hanoi.s and then python lc5500-sim.py hanoi.bin to run your code. If you cannot do this with your submission, then you have done something wrong.
5 Appendix A: LC-5500 Instruction Set Architecture
The LC-5500 is a simple, yet capable computer architecture. The LC-5500 combines attributes of both ARM and the LC-2200 ISA defined in the Ramachandran & Leahy textbook for CS 2200.
The LC-5500 is a word-addressable, 32-bit computer. All addresses refer to words, i.e. the first word (four bytes) in memory occupies address 0x0, the second word, 0x1, etc.
All memory addresses are truncated to 16 bits on access, discarding the 16 most significant bits if the address was stored in a 32-bit register. This provides roughly 64 KB of addressable memory.
5.1 Registers
The LC-5500 has 16 general-purpose registers. While there are no hardware-enforced restraints on the uses of these registers, your code is expected to follow the conventions outlined below.
Table 1: Registers and their Uses
Register Number Name Use Callee Save?
0 $zero Always Zero NA
1 $at Assembler/Target Address NA
2 $v0 Return Value No
3 $a0 Argument 1 No
4 $a1 Argument 2 No
5 $a2 Argument 3 No
6 $t0 Temporary Variable No
7 $t1 Temporary Variable No
8 $t2 Temporary Variable No
9 $s0 Saved Register Yes
10 $s1 Saved Register Yes
11 $s2 Saved Register Yes
12 $k0 Reserved for OS and Traps NA
13 $sp Stack Pointer No
14 $fp Frame Pointer Yes
15 $ra Return Address No
1. Register 0 is always read as zero. Any values written to it are discarded. Note: for the purposes of this project, you must implement the zero register. Regardless of what is written to this register, it should always output zero.
3. Register 2 is where you should store any returned value from a subroutine call.
4. Registers 3 – 5 are used to store function/subroutine arguments. Note: registers 2 through 8 should be placed on the stack if the caller wants to retain those values. These registers are fair game for the callee (subroutine) to trash.
5. Registers 6 – 8 are designated for temporary variables. The caller must save these registers if they want these values to be retained.
7. Register 12 is reserved for handling interrupts. While it should be implemented, it otherwise will not have any special use on this assignment.
8. Register 13 is the everchanging top of the stack stack; it keeps track of the top of the activation record for a subroutine.
9. Register 14 is the anchor point of the activation frame. It is used to point to the first address on the activation record for the currently executing process.
10. Register 15 is used to store the address a subroutine should return to when it is finished executing.
5.2 Instruction Overview
The LC-5500 supports a variety of instruction forms, only a few of which we will use for this project. The instructions we will implement in this project are summarized below.
Table 2: LC-5500 Instruction Set
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0000 DR SR1 unused SR2
0001 DR SR1 unused SR2
0010 DR SR immval20
0011 DR BaseR offset20
0100 SR BaseR offset20
0101 unused offset20
0110 RA AT unused
0111 unused
1000 SR1 SR2 PCoffset20
1001 DR unused PCoffset20
1010 SR $SP offset20 (-1)
1011 DR $SP offset20 (+1)
ADD
NAND
ADDI
LW
SW
BR
JALR
HALT
BLT
LEA
PUSH
POP
5.3 Detailed Instruction Reference
5.3.1 ADD
Assembler Syntax
ADD DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0000 DR SR1 unused SR2
Operation
DR = SR1 + SR2;
Description
The ADD instruction obtains the first source operand from the SR1 register. The second source operand is obtained from the SR2 register. The second operand is added to the first source operand, and the result is stored in DR.
5.3.2 NAND
Assembler Syntax
NAND DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0001 DR SR1 unused SR2
Operation
DR = ~(SR1 & SR2);
Description
The NAND instruction performs a logical NAND (AND NOT) on the source operands obtained from SR1 and SR2. The result is stored in DR.
HINT: A logical NOT can be achieved by performing a NAND with both source operands the same.
For instance,
NAND DR, SR1, SR1

…achieves the following logical operation: DR ← SR1.
5.3.3 ADDI
Assembler Syntax
ADDI DR, SR, immval20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0010 DR SR immval20
Operation
DR = SR + SEXT(immval20);
Description
The ADDI instruction obtains the first source operand from the SR register. The second source operand is obtained by sign-extending the immval20 field to 32 bits. The resulting operand is added to the first source operand, and the result is stored in DR.
5.3.4 LW
Assembler Syntax
LW DR, offset20(BaseR)
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0011 DR BaseR offset20
Operation
DR = MEM[BaseR + SEXT(offset20)];
Description
An address is computed by sign-extending bits [19:0] to 32 bits and then adding this result to the contents of the register specified by bits [23:20]. The 32-bit word at this address is loaded into DR.

5.3.5 SW
SW SR, offset20(BaseR)
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0100 SR BaseR offset20
Operation
MEM[BaseR + SEXT(offset20)] = SR;
Description
An address is computed by sign-extending bits [19:0] to 32 bits and then adding this result to the contents of the register specified by bits [23:20]. The 32-bit word obtained from register SR is then stored at this address.
5.3.6 BR
Assembler Syntax
BR offset20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0101 unused offset20
Operation
PC = incrementedPC + offset20
Description
A branch is unconditionally taken. The PC will be set to the sum of the incremented PC (since we have already undergone fetch) and the sign-extended offset[19:0].
5.3.7 JALR
JALR RA, AT
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0110 RA AT unused
Operation
RA = PC;
PC = AT;
Description
First, the incremented PC (address of the instruction + 1) is stored into register RA. Next, the PC is loaded with the value of register AT, and the computer resumes execution at the new PC.
5.3.8 HALT
Assembler Syntax
HALT
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0111 unused
Description
The machine is brought to a halt and executes no further instructions.
5.3.9 BLT
Assembler Syntax
BLT SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1000 SR1 SR2 PCoffset20
Operation
if (SR1 < SR2) {
PC = incrementedPC + SEXT(PCoffset20)
}
Description
This is a conditional branch that will be taken only if the value of the register SR1 is less that then value of the SR2 register. The PC will be set to the sum of the incremented PC (since we have already undergone fetch) and the sign-extended offset[19:0].
5.3.10 LEA
LEA DR, label
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1001 DR unused PCoffset20
Operation
DR = PC + SEXT(PCoffset20);
Description
An address is computed by sign-extending bits [19:0] to 32 bits and adding this result to the incremented PC (address of instruction + 1). It then stores the computed address into register DR.
5.3.11 PUSH
Assembler Syntax
PUSH SR
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1010 SR $SP offset20 (-1)
Operation
X = SR
$SP = $SP – 1 MEM[$SP] = X
Description
Decrements the stack pointer and stores the value of the source register SR at the memory location of the stack pointer. Note: the operation of the decrement must happen before the store in order to ensure that the stack pointer always points to the top of the stack, but be sure to take special care when pushing the stack pointer itself.

5.3.12 POP
Assembler Syntax
POP DR
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1010 DR $SP offset20 (+1)
Operation
X = MEM[$SP]
$SP = $SP + 1
DR = X
Description
Loads the value from memory pointed to by the stack pointer into DR and increments the value of the stack pointer. Note: the operation of the increment must happen after the load in order to ensure that the stack pointer always points to the top of the stack.