辅导案例-CS 2506

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CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 1
The Assignment

Recall that an assembler translates code written in mnemonic form in assembly language into machine code.

You will implement an assembler that supports a subset of the MIPS32 assembly language (specified below). The
assembler will be implemented in C and executed on Linux. Your assembler will take a file written in that subset of the
MIPS32 assembly language, and write an output file containing the corresponding MIPS32 machine code (in text format).


Supported MIPS32 Assembly Language

The subset of the MIPS assembly language that you need to implement is defined below. The following conventions are
observed in this section:

• The notation (m:n) refers to a range of bits in the value to which it is applied. For example, the expression

(PC+4)(31:28)

refers to the high 4 bits of the address PC+4.
• imm16 refers to a 16-bit immediate value; no assembly instruction will use a longer immediate
• offset refers to a literal applied to an address in a register; e.g., offset(rs); offsets will be signed 16-bit
values
• label fields map to addresses, which will always be 16-bit values if you follow the instructions
• sa refers to the shift amount field of an R-format instruction; shift amounts will be nonnegative 5-bit values
• target refers to a 26-bit word address for an instruction; usually a symbolic notation
• Sign-extension of immediates is assumed where necessary and not indicated in the notation.
• C-like notation is used in the comments for arithmetic and logical bit-shifts: >>a , <>l
• The C ternary operator is used for selection: condition ? if-true : if-false
• Concatenation of bit-sequences is denoted by ||.
• A few of the specified instructions are actually pseudo-instructions. That means your assembler must replace each
of them with a sequence of one or more other instructions; see the comments for details.

You will find the MIPS32 Architecture Volume 2: The MIPS32 Instruction Set to be an essential reference for machine
instruction formats and opcodes, and even information about the execution of the instructions. See the Resources page on
the course website.


MIPS32 assembly .data section:

.word This is used to hold one or more 32 bit quantities, initialized with given values. For example:

var1: .word 15 # creates one 32 bit integer called var1 and initializes
# it to 15.

array1: .word 2, 3, 4, 5, 6, 7 # creates an array of 6 32-bit integers,
# initialized as indicated (in order).

array2: .word 2:10 # creates an array of 10 32 bit integers and initializes
# element of the array to the value 2.

.asciiz This is used to hold a NULL (0) terminated ASCII string. For example:

hello_w: .asciiz "hello world" # this declaration creates a NULL
# terminated string with 12 characters
# including the terminating 0 byte


CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 2
MIPS32 assembly .text section:

Your assembler must support translation of all of the following MIPS32 assembly instructions. You should consult the
MIPS32 Instruction Set reference for opcodes and machine instruction format information. That said, we will evaluate your
output by comparing it to the instruction specifications in the MIPS32 Instruction Set manual.

Your assembler must support the following basic load and store instructions:

lw rt, offset(rs) # Transfers a word from memory to a register
# GPR[rt] <-- Mem[GPR[rs] + offset]

sw rt, offset(rs) # Transfers a word from a register to memory
# Mem[GPR[rs] + offset] <-- GPR[rt]

lui rt, imm16 # Loads a constant into the upper half of a register
# GPR[rt] <-- imm16 || 0x0000

Note that the constant offset in the load and store instructions (lw and sw) may be positive or negative.

Your assembler must support the following basic arithmetic/logical instructions:

add rd, rs, rt # signed addition of integers; overflow detection
# GPR[rd] <-- GPR[rs] + GPR[rt]

addi rt, rs, imm16 # signed addition with 16-bit immediate;
# overflow detection
# GPR[rt] <-- GPR[rs] + imm16

addu rd, rs, rt # unsigned addition with 16-bit immediate;
# no overflow detection
# GPR[rt] <-- GPR[rs] + GPR[rt]

addiu rt, rs, imm16 # unsigned addition with 16-bit immediate;
# no overflow detection
# GPR[rt] <-- GPR[rs] + imm16

and rd, rs, rt # bitwise logical AND
# GPR[rd] <-- GPR[rs] AND GPR[rt]

andi rt, rs, imm16 # bitwise logical AND with 16-bit immediate
# GPR[rd] <-- GPR[rs] AND imm16

mul rd, rs, rt # signed multiplication of integers;
# no overflow detection
# GPR[rd] <-- (GPR[rs] * GPR[rt])(31:0)

nop # no operation
# executed as: sll $zero, $zero, 0

nor rd, rs, rt # bitwise logical NOR
# GPR[rd] <-- !(GPR[rs] OR GPR[rt])

sll rd, rt, sa # logical shift left a fixed number of bits
# GPR[rd] <-- GPR[rs] <
slt rd, rs, rt # set register to result of comparison
# GPR[rd] <-- (GPR[rs] < GPR[rt] ? 0 : 1)

slti rt, rs, imm16 # set register to result of comparison
# GPR[rt] <-- (GPR[rs] < imm16 ? 0 : 1)

sra rd, rt, sa # arithmetic shift right a fixed number of bits
# GPR[rd] <-- GPR[rt] >>a sa
CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 3

srav rd, rt, rs # arithmetic shift right a variable number of bits
# GPR[rd] <-- GPR[rt] >>a GPR[rs]

sub rd, rs, rt # signed subtraction of integers
# GPR[rd] <-- GPR[rs] - GPR[rt]

Your assembler must support the following basic control-of-flow instructions:

beq rs, rt, offset # conditional branch if rs == rt
# PC <-- (GPR[rs] == GPR[rt] ? PC + 4 + offset < # : PC + 4)

blez rs, offset # conditional branch if rs <= 0
# PC <-- (GPR[rs] <= 0 ? PC + 4 + offset < # : PC + 4)

bgtz rs, offset # conditional branch if rs > 0
# PC <-- (GPR[rs] > 0 ? PC + 4 + offset < # : PC + 4)

bne rs, rt, offset # conditional branch if rs != rt
# PC <-- (GPR[rs] != GPR[rt] ? PC + 4 + offset < # : PC + 4)

j target # unconditional branch
# PC <-- ( (PC+4)(31:28) || (target <
syscall # invoke exception handler, which examines $v0
# to determine appropriate action; if it returns,
# returns to the succeeding instruction; see the
# MIPS32 Instruction Reference for format

Your assembler must support the following pseudo-instructions:

move rd, rs # copy contents of GPR[rs] to GPR[rt]
# GPR[rd] = GPR[rs]
# pseudo-translation:
# addu rd, zero, rs


blt rs, rt, offset # conditional branch if rs < rt
# PC <-- (GPR[rs] < GPR[rt] ? PC + 4 + offset < # : PC + 4)
# pseudo-translation:
# slt at, rs, rt
# bne at, zero, offset

la rt, label # load address label to register
# GPR[rd] <-- label
# pseudo-translation for 16-bit label:
# addi rt, $zero, label

li rt, imm16 # load 16-bit immediate to register
# GPR[rd] <-- imm
# pseudo-translation:
# addiu rt, $zero, imm16

lw rt, label # load word at address label to register
# GPR[rd] <-- Mem[label]
# pseudo-translation:
# lw rt, label[15:0]($zero)


CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 4
MIPS32 assembly format constraints:

The assembly programs will satisfy the following constraints:

• Labels will begin in the first column of a line, and will be no more than 32 characters long. Labels are restricted to
alphanumeric characters and underscores, and are always followed immediately by a colon character (':').
• Labels in the .text segment will always be on a line by themselves.
• Labels in the .data segment will always occur on the same line as the specification of the variable being defined.
• Labels are case-sensitive; that actually makes your task a bit simpler.
• MIPS instructions do not begin in a fixed column; they are preceded by an arbitrary amount of whitespace
(possibly none).
• Blank lines may occur anywhere; a blank line will always contain only a newline character.
• Whitespace will consist of spaces, tab characters, or a mixture of the two. Your parsing logic must handle that.
• Registers will be referred to by symbolic names ($zero, $t5) rather than by register number.
• Instruction mnemonics and register names will use lower-case characters.
• Assembly source files will always be in UNIX format.

You must be sure to test your implementation with all the posted test files; that way you should avoid any unfortunate
surprises when we test your implementation.


Input

The input files will be MIPS assembly programs in ASCII text. The assembly programs will be syntactically correct,
compatible with the MIPS32 Instruction Set manual, and restricted to the subset of the MIPS32 instruction set defined
above. Example programs will be available from the course website.

Each line in the input assembly file will either contain an assembly instruction, a section header directive (such as .data)
or a label (a jump or branch target). The maximum length of a line is 256 bytes.

Your input file may also contain comments. Any text after a '#' symbol is a comment and should be discarded by your
assembler. Section header directives, such as .data and .text will be in a line by themselves. Similarly, labels (such as
loop:) will be on a line by themselves. The input assembly file will contain one data section, followed by one text
section.

Your assembler can be invoked in either of the following ways:

assemble
assemble –symbols -symbols

The specified input file must already exist; if not, your program should exit gracefully with an error message to the console
window. The specified output file may or may not already exist; if it does exist, the contents should be overwritten.


Output

Output when invoked as: assemble

Your assembler will resolve all references to branch targets in the .text section and variables in the .data section and
convert the instructions in the .text section into machine code.

To convert an instruction into machine code follow the instruction format rules specified in the class textbook. For each
format (R-format, I-format or J-format), you should determine the opcode that corresponds to instruction, the values for the
register fields and any optional fields such as the function code and shift amount fields for arithmetic instructions (R-
format) and immediate values for I-format instructions.

CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 5
The output machine code should be saved to the output file specified in the command line. The output file should contain
the machine code corresponding to instructions from the .text section followed by a blank line followed by variables
from the .data section in human readable binary format (0s and 1s). For example, to represent the decimal number 40 in
16-bit binary you would write 0000000000101000, and to represent the decimal number -40 in 16-bit binary you would
write 1111111111011000.

The output file is a text file, not a binary file; that’s a concession to the need to evaluate your results.

Your output file should match the machine code file posted with the grading harness. A sample showing the assembler’s
translation of the adder.asm program is given at the end of this specification.

Output when invoked as: assemble –symbols

Your assembler will write (to the specified output file) a well-formatted table, listing every symbolic name used in the
MIPS32 assembly code and the address that corresponds to that label. Addresses will be written in hex. Note: when
invoked this way, your assembler will not write any other output.

We will make the following assumptions about addresses and program segments:

• The base address for the text segment is 0x00000000, so that’s the address of the first machine instruction.
• The base address of the data segment is 0x00002000, so that’s the address of the first thing declared in the data
segment.

The second fact above implies that the text segment cannot be longer than 8 KiB or 2048 machine instructions. You don’t
need to do anything special about that fact.



CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 6
Sample Assembler Input

# adder.asm
#
# The following program computes the sum of all elements in an array.
# The program then prints the sum of all the entries in the array.

.data
message: .asciiz "The sum of the numbers in the array is: "
values: .word 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 # array of 10 words
num_values: .word 10 # size of array
sum: .word 0 # running total

.text
main:
la $s0, values # load address of the array;
# $s0 will point to the current element
la $s1, num_values # load address of num_values
lw $s1, 0($s1) # load num_values
sll $s1, $s1, 2 # compute size of array in bytes
add $s1, $s0, $s1 # compute one-past-end address
li $s2, 0 # set running total to 0

j chk # check for empty array
loop:
lw $s3, 0($s0) # fetch current data element
add $s2, $s2, $s3 # update running total
addi $s0, $s0, 4 # step pointer to next array element
chk:
sub $t0, $s0, $s1 # compute distance beyond one-past-end
blez $t0, loop # if ( distance <= 0 ) then loop

# done processing array elements; store total to memory
la $s4, sum # get address for sum
sw $s2, 0($s4) # write sum to memory

# report results to user
li $v0, 4 # system call to print string
la $a0, message # load address of message into arg register
syscall # make call

li $v0, 1 # system call to print an integer
and $a0, $s2, $s2 # load value to print into arg register
syscall # make call

li $v0, 10 # system call to terminate program
syscall # make call



CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 7
Sample Assembler Output

Here’s my assembler’s output when invoked as: assemble adder.asm adder.o:


00100000000100000010000000101100 la
00100000000100010010000001010100 la
10001110001100010000000000000000 lw
00000000000100011000100010000000 sll
00000010000100011000100000100000 add
00100100000100100000000000000000 li
00001000000000000000000000001010 j
10001110000100110000000000000000 lw
00000010010100111001000000100000 add
00100010000100000000000000000100 addi
00000010000100010100000000100010 sub
00011001000000001111111111111011 blez
00100000000101000010000001011000 la
10101110100100100000000000000000 sw
00100100000000100000000000000100 li
00100000000001000010000000000000 la
00000000000000000000000000001100 syscall
00100100000000100000000000000001 li
00000010010100100010000000100100 and
00000000000000000000000000001100 syscall
00100100000000100000000000001010 li
00000000000000000000000000001100 syscall

01010100011010000110010100100000
01110011011101010110110100100000
01101111011001100010000001110100
01101000011001010010000001101110
01110101011011010110001001100101
01110010011100110010000001101001
01101110001000000111010001101000
01100101001000000110000101110010
01110010011000010111100100100000
01101001011100110011101000100000
00000000000000000000000000000000
00000000000000000000000000000010
00000000000000000000000000000011
00000000000000000000000000000101
00000000000000000000000000000111
00000000000000000000000000001011
00000000000000000000000000001101
00000000000000000000000000010001
00000000000000000000000000010011
00000000000000000000000000010111
00000000000000000000000000011101
00000000000000000000000000001010
00000000000000000000000000000000

values
num_values
sum
text segment for adder.o
data segment for adder.o
message +
terminator +
padding
CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 8
Data segment notes:

The character string variable message is stored as a sequence of one-byte ASCII codes, with the characters in ascending
order by address.

The next value in the data segment is a 32-byte integer, which must be aligned on an address that is a multiple of 4;
therefore, we must add "padding" NULL bytes before storing the bytes of the integer. The integer variable values is
stored as a sequence of 32-bit values, which are the 2's complement representations of the values shown in the assembly
code. In the data segment display above, the first value, 2, is displayed in big-endian byte order:

00000000 00000000 00000000 00000010
Low address High address

The value of num_values is 10, which is expressed in hex as 0x0000000A.

00000000 00000000 00000000 00001010

You must be sure that you write the bytes of integers in big-endian order as well.


Handling of strings in the data segment:

For example, suppose we have the data segment:

.data
S1: .asciiz "abc"
I1: .word 1
S4: .asciiz "CDEFG"
I4: .word 4

The corresponding data segment representation should be:

01100001011000100110001100000000 S1: 'a' 'b' 'c' '\0'
00000000000000000000000000000001 I1: 0x00000001
01000011010001000100010101000110 S2: 'C' 'D' 'E' 'F'
01000111000000000000000000000000 'G' '\0' padding bytes
00000000000000000000000000000100 I2: 0x00000004

The ASCII codes for the characters in a string are stored in front-to-back order (the opposite of the usual MIPS
convention). Since MIPS requires a 32-bit integer to be aligned to an address that is a multiple of 4, we must insert padding
bytes after the string variable S2. We will place those padding bytes as shown above, after the last byte of the string
variable ('\0').


Output for the -symbols option:

Here’s my assembler’s output when invoked as: assemble -symbols adder.asm adder.sym:

0x00002000 message
0x0000202C values
0x00002054 num_values
0x00002058 sum
0x00000000 main
0x0000001C loop
0x00000028 chk

The order in which you display the symbols is up to you. My implementation writes them in the order they occur in the
source file, but that is not a requirement.
CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 9
How can I verify my output or test my code?

Download the posted test/grading tar file and unpack it. This will provide you with a collection of MIPS assembly (.asm)
files and corresponding assembled object files (.o), placed in a subdirectory testData. For example, if you’ve compiled
your assembler, you could run it on the first test case by using the command:

assemble ./testData/test01.asm myobj01.o

You can then compare your assembler’s output to the reference output by using the supplied compare utility:

compare 1 ./testData/test01.o myobj01.o

This will generate detailed output showing any mismatches, and a score. We will use exactly this approach in evaluating
the correctness of your submission. The first parameter is just a dummy one that has no significance until it is used by the
grading code.



CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 10
Valgrind

Valgrind is a tool for detecting certain memory-related errors, including out of bounds accessed to dynamically-allocated
arrays and memory leaks (failure to deallocate memory that was allocated dynamically). A short introduction to Valgrind is
posted on the Resources page, and an extensive manual is available at the Valgrind project site (www.valgrind.org).

For best results, you should compile your C program with a debugging switch (-g or –ggdb3); this allows Valgrind to
provide more precise information about the sources of errors it detects. For example, I ran my solution for a related project,
with one of the test cases, on Valgrind:

[wdm@centosvm C3]$ valgrind --leak-check=full --show-leak-kinds=all --log-file=vlog.txt
--track-origins=yes -v disassem C3TestFiles/ref07.o stu_ref07.asm
==27447== Memcheck, a memory error detector
==27447== Copyright (C) 2002-2013, and GNU GPL'd, by Julian Seward et al.
==27447== Using Valgrind-3.10.0 and LibVEX; rerun with -h for copyright info
==27447== Command: disassem C3TestFiles/ref07.o stu_ref07.asm
==27447== Parent PID: 3523
==27447==
==27447==
==27447== HEAP SUMMARY:
==27447== in use at exit: 0 bytes in 0 blocks
==27447== total heap usage: 226 allocs, 226 frees, 6,720 bytes allocated
==27447==
==27447== All heap blocks were freed -- no leaks are possible
==27447==
==27447== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 2 from 2)
==27447==
==27447== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 2 from 2)

And, I got good news... there were no detected memory-related issues with my code.

On the other hand, I ran Valgrind on a different solution to the same project, and the results were less satisfactory:

[wdm@centosvm temp]$ valgrind --leak-check=full ./driver t01.o t01.txt -rand
. . .
==7962== HEAP SUMMARY:
==7962== in use at exit: 4,842 bytes in 331 blocks
==7962== total heap usage: 331 allocs, 0 frees, 4,842 bytes allocated
==7962==
==7962== Searching for pointers to 331 not-freed blocks
==7962== Checked 111,584 bytes
==7962==
==7962== 7 bytes in 1 blocks are definitely lost in loss record 1 of 26
==7962== at 0x4C2B974: calloc (in /usr/lib64/valgrind/vgpreload_memcheck-amd64-linux.so)
==7962== by 0x4020B3: parseJTypeInstruction (ParseInstructions.c:178)
==7962== by 0x400F51: main (Disassembler.c:137)
. . .
==7962== LEAK SUMMARY:
==7962== definitely lost: 3,116 bytes in 233 blocks
==7962== indirectly lost: 590 bytes in 96 blocks
==7962== possibly lost: 0 bytes in 0 blocks
==7962== still reachable: 1,136 bytes in 2 blocks
==7962== suppressed: 0 bytes in 0 blocks
==7962==
==7962== ERROR SUMMARY: 116 errors from 34 contexts (suppressed: 2 from 2)

It's worth noting that Valgrind not only detects the occurrence of memory leaks, but it also pinpoints where the leaked
memory was allocated in the code. That makes it much easier to track down the logical errors that lead to the leaks.

Valgrind can also detect computations that use uninitialized variables, and invalid reads/writes (to memory locations that lie
outside the user's requested dynamic allocation).

Do note that Valgrind has its limitations. False positives are uncommon, as are false negatives, but both are possible. More
importantly, Valgrind depends on using its own internal memory allocator in order to detect memory leaks and out-of-
bounds memory accesses. Valgrind is not particularly good at detecting errors related to statically-allocated memory.

CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 11
Milestones
In order to assess your progress, there will be two milestones for the project. Each of these will require that you submit a
partial solution that achieves specified functionality. Each milestone will be evaluated by using a scripted testing
environment, which will be posted on the course website at least two weeks before the corresponding milestone is due. We
will not perform any Valgrind-based evaluation of the milestones, although the test harness may include relevant output.

Your scores on the milestones will constitute 6% and 9%, respectively, of your final score on the project.

Milestone 1

The first milestone will be due approximately two weeks before the final project deadline. Your submission must support
translation of a MIPS assembly program that consists of a .text segment including the following instructions:

add rd, rs, rt # signed addition of integers; overflow detection
# GPR[rd] <-- GPR[rs] + GPR[rt]

addi rt, rs, imm16 # signed addition with 16-bit immediate;
# overflow detection
# GPR[rt] <-- GPR[rs] + imm16

nor rd, rs, rt # bitwise logical NOR
# GPR[rd] <-- !(GPR[rs] OR GPR[rt])

slti rt, rs, imm16 # set register to result of comparison
# GPR[rt] <-- (GPR[rs] < imm16 ? 0 : 1)

syscall # invoke exception handler, which examines $v0
# to determine appropriate action; if it returns,
# returns to the succeeding instruction; see the
# MIPS32 Instruction Reference for format

Your submission for milestone 1 must support references to the $s* and $v0 registers. The test files for this milestone
will not include a .data segment, and there will be no symbolic labels in the .text segment.

Milestone 2

The second milestone will be due approximately one week before the final submission. Your submission for this milestone
must support translation of a MIPS assembly program that includes both a .data and a .text segment. The .data
segment may include:

.word This is used to hold one or more 32 bit quantities, initialized with given values. For example:

var1: .word 15 # creates one 32 bit integer called var1 and initializes
# it to 15.

array2: .word 2:10 # creates an array of 10 32 bit integers and initializes
# all the elements of the array to the value 2.

The .text segment may include any of the instructions from the first milestone, and also:

lw rt, offset(rs) # Transfers a word from memory to a register.
# GPR[rt] <-- Mem[GPR[rs] + offset]

la rd, label # load address label to register
# GPR[rd] <-- label
# pseudo-translation for 16-bit label:
# addi rd, $zero, label


CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 12
mul rd, rs, rt # signed multiplication of integers;
# no overflow detection
# GPR[rd] <-- (GPR[rs] * GPR[rt])(31:0)

nor rd, rs, rt # bitwise logical NOR
# GPR[rd] <-- !(GPR[rs] OR GPR[rt])

beq rs, rt, offset # conditional branch if rs == rt
# PC <-- (GPR[rs] == GPR[rt] ? PC + 4 + offset < # : PC + 4)

bne rs, rt, offset # conditional branch if rs != rt
# PC <-- (GPR[rs] != GPR[rt] ? PC + 4 + offset < # : PC + 4)

Your submission for milestone 2 must deal with all requirements for milestone 1, support the $s*, $v0 and $zero
registers, and symbolic labels in both the .text and .data segments.

NO LATE SUBMISSIONS WILL BE GRADED FOR EITHER MILESTONE!


Extra Credit
For 10% extra credit, implement your assembler so that it will handle a MIPS32 assembly program with a single data
segment and a single text segment, in either order. The feature must operate automatically, without any extra command-
line switches or recompilation. This will be evaluated by performing a single test, and there will be no partial credit for
the feature.


What should I turn in, and how?
For both the milestone and the final submissions, create an uncompressed tar file containing:

• All the .c and .h files which are necessary in order to build your assembler.
• A GNU makefile named "makefile". The command "make assembler" should build an executable
named "assemble". The makefile may include additional targets as you see fit.
• A readme.txt file if there’s anything you want to tell us regarding your implementation. For example, if there
are certain things that would cause your assembler to fail (e.g., it doesn’t handle la instructions), telling us that
may result in a more satisfactory evaluation of your assembler.
• A pledge.txt file containing the pledge statement from the course website.
• Nothing else. Do not include object files or an executable. We will compile your source code.

Late submissions of the final project will be penalized at a rate of 10% per day until the final submission deadline.

A flat tar file is one that includes no directory structure. In this case, you can be sure you've got it right by performing a
very simple exercise. Unpack the posted test harness, and follow the instructions in the readme.txt file. If the two shell
scripts fail to build an executable, and test it, then there's something wrong with your tar file (or your C code).

You can also tell the difference by simply doing a table-of-contents listing of your tar file. The listing of a flat tar file will
not show any path information. The one on the left below is not flat; the one on the right is flat:

CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 13

Linux > tar tf C3Test.tar
C3TestFiles/
C3TestFiles/mref01.asm
C3TestFiles/mref01.o
C3TestFiles/mref02.asm
C3TestFiles/mref02.o
C3TestFiles/mref03.asm
C3TestFiles/mref03.o
compare
prepC3.sh
testC3.sh
Linux > tar tf C3Source.tar
assembler.c
Instruction.c
IWrapper.c
MIParser.c
ParseResult.c
Registers.c
SymbolTable.c
Instruction.h
MIParser.h
ParseResult.h
Registers.h
SymbolTable.h
SystemConstants.h
makefile
pledge.txt


Some General Coding Requirements

Your solution will be compiled by a test/grading harness that will be supplied along with this specification.

You are required* to implement your solution in logically-cohesive modules (paired .h and .c files), where each module
encapsulates the code and data necessary to perform one logically-necessary task. For example, a module might
encapsulate the task of mapping register numbers to symbolic names, or the task of mapping mnemonics to opcodes, etc.
There are many reasonable ways to organize the code for a system as large as this; my solution employs about a dozen
modules, and involves more than 2300 lines of C code.

The TAs are instructed that they are not required to provide help with solutions that are not properly organized into
different files, or with solutions that are not adequately commented.

We will require* your solution to achieve a "clean" run on valgrind. A clean run should report the same number of
allocations and frees, that zero heap bytes were in use when the program terminated, that there were no invalid reads or
writes, and that there were no issues with uninitialized data. We will not be concerned about suppressed errors reported by
valgrind. See the discussion of valgrind below, and the introduction to valgrind that's posted on the course
website.

In line with the requirement above, we require that you make sensible use of dynamic memory allocation in your solution.
Just how and where you allocate objects dynamically is up to you, but failure to use dynamic allocation sufficiently will
result in a deduction.

Finally, this is not a requirement, but you are strongly advised to use calloc() when you allocate dynamically, rather
than malloc(). This will guarantee your dynamically-allocated memory is zeroed when it's allocated, and that may help
prevent certain errors.

* "Required" here means that this will be scored by a human being after your solution has been autograded. The automated
evaluation will certainly not adjust your score for these things. Failure to satisfy these requirements will result in
deductions from your autograding score; the potential size of those deductions may not be specified in advance (but you
will not be happy with them).



CS 2506 Computer Organization II C03: Simple MIPS Assembler
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Grading
The evaluation of your solution will be based entirely on its ability to correctly translate programs using the specified
MIPS32 assembly subset to MIPS32 machine code. That is somewhat unfortunate, since there are many other issues we
would like to consider, such as the quality of your design, your internal documentation, and so forth. However, we do not
have sufficient staff to consider those things fairly, and therefore we will not consider them at all.

At least two weeks before the due date for each milestone, we will release a tar file containing a testing harness (test shell
scripts and test cases). You can use this tar file to evaluate your milestone submission, in advance, in precisely the same
way we will evaluate it. We are posting the test harness as an aid in your testing, but also so that you can verify that you
are packaging your submission according to the requirements given above. Submissions that do not meet the requirements
typically receive extremely low scores.

The testing of your final assembler submission will simply add test cases to cover the full range of specified MIPS32
instructions and data declarations, and to evaluate any extra-credit features that may be specified for this assignment. We
will release an updated test harness for the final submissions at least two weeks before the final deadline.

Our testing of your milestone and final assembler submissions will be performed using the test harness files we will make
available to you. We expect you to use each test harness to validate your solution.

We will not offer any accommodations for submissions that do not work properly with the corresponding supplied
test harness.


Test Environment

Your assembler will be tested on the rlogin cluster or the equivalent, running 64-bit CentOS 7 and gcc 4.8. There are many
of you, and few of us. Therefore, we will not test your assembler on any other environment. So, be sure that you compile
and test it there before you submit it. Be warned in particular, if you use OS X, that the version of gcc available there has
been modified for Apple-specific reasons, and that past students have encountered significant differences between that
version and the one running on Linux systems.


Maximizing Your Results

Ideally you will produce a fully complete and correct solution. If not, there are some things you can do that are likely to
improve your score:

• Make sure your assembler submission works properly with the posted test harness (described above). If it does
not, we will almost certainly not be able to evaluate your submission and you are likely to receive a score of 0.
• Make sure your assembler does not crash on any valid input, even if it cannot produce the correct results. If you
ensure that your assembler processes all the posted test files, it is extremely unlikely it will encounter anything in
our test data that would cause it to crash. On the other hand, if your assembler does crash on any of the posted test
files, it will certainly do so during our testing. We will not invest time or effort in diagnosing the cause of such a
crash during our testing. It’s your responsibility to make sure we don’t encounter such crashes.
• If there is a MIPS32 instruction or data declaration that your solution cannot handle, document that in the
readme.txt file you will include in your submission.
• If there is a MIPS32 instruction or data declaration that your solution cannot handle, make sure that it still
produces the correct number of lines of output, since we will automate much of the checking we do. In particular,
if your assembler encounters a MIPS32 instruction it cannot handle, write a sequence of 32 asterisk characters
('*') in place of the correct machine representation (or multiple lines for some pseudo-instructions). Doing this
will not give you credit for correctly translating that instruction, but this will make it more likely that we correctly
evaluate the following parts of your translation.



CS 2506 Computer Organization II C03: Simple MIPS Assembler
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Pledge:

Each of your program submissions must be pledged to conform to the Honor Code requirements for this course.
Specifically, you must include a file, named pledge.txt, containing the following pledge statement in the submitted tar
file:

// On my honor:
//
// - I have not discussed the C language code in my program with
// anyone other than my instructor or the teaching assistants
// assigned to this course.
//
// - I have not used C language code obtained from another student,
// or any other unauthorized source, either modified or unmodified.
//
// - If any C language code or documentation used in my program
// was obtained from another source, such as a text book or course
// notes, that has been clearly noted with a proper citation in
// the comments of my program.
//
//

Failure to include the pledge statement may result in your submission being ignored.


Credits

The original formulation of this project was created by Dr Srinidhi Vadarajan, who was then a member of the Dept of
Computer Science at Virginia Tech. His sources of inspiration for this project are lost in the mists of time.

The current modification was produced by William D McQuain, as a member of the Dept of Computer Science at Virginia
Tech. Any errors, ambiguities and omissions should be attributed to him.



CS 2506 Computer Organization II C03: Simple MIPS Assembler
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Advice

The following observations are purely advisory, but are based on my experience, including that of implementing a solution
to this assignment. These are advice, not requirements.

First, and most basic, analyze what your assembler must do and design a sensible, logical framework for making those
things happen. There are fundamental decisions you must make early in the process of development. For example, you
could represent the machine instructions in a number of ways as you build them. They can be represented as arrays of
individual bits (which could be integers or characters), or they can be represented in binary format, which would be the
expected format for a "real" assembler's final output. I am not convinced that either of those approaches is inherently
better, or that there are not reasonable alternatives. But, this decision has ramifications that will propagate throughout your
implementation.

It helps to consider how you would carry out the translation from assembly code to machine instructions by hand. If you do
not understand that, you are trying to write a program that will do something you do not understand, and your chances of
success are reduced to sheer dumb luck.

Second, and also basic, practice incremental development! This is a sizeable program, especially so if it's done properly.
My solution, including comments, runs something over 2300 lines of code. It takes quite a bit of work before you have
enough working code to test on full input files, but unit testing is extremely valuable.

Record your design decisions in some way; a simple text file is often useful for tracking your deliberations, the alternatives
you considered, and the conclusions you reached. That information is invaluable as your implementation becomes more
complete, and hence more complex, and you are attempting to extend it to incorporate additional features.

Write useful comments in your code, as you go. Leave notes to yourself about things that still need to be done, or that you
are currently handling in a clumsy manner.

A preprocessing phase is helpful; for example, it gives you a chance to filter out comments, trim whitespace, and gather
various pieces of information. Do not try to do everything in one pass. Compilers and assemblers frequently produce a
number of intermediate files and/or in-memory structures, recording the results of different phases of execution.

Consider how you would carry out the translation of a MIPS32 assembly program to machine code if you were doing it
manually. If you don't understand how to do it by hand, you cannot write a program to do it!

Take advantage of tools. You should already have a working knowledge of gdb. Use it! The debugger is invaluable when
pinning down the location of segfaults; but it is also useful for tracking down lesser issues if you make good use of
breakpoints and watchpoints. Some memory-related errors yield mysterious behavior, and confusing runtime error reports.
That's especially true when you have written past the end of a dynamically-allocated array and corrupted the heap. This
sort of error can often be diagnosed by using valgrind.

Enumerated types are extremely useful for representing various kinds of information, especially about type attributes of
structured variables. For example, if implementing a GIS system, we might find the following type useful:

// GData.h
enum _FeatureType {CITY, RIVER, MOUNTAIN, BUILDING, . . . , ISLAND};
typedef enum _FeatureType FeatureType;
...
struct _GData {
char* Name;
char* State;
...
FeatureType FType;
uint16_t Elevation;
};
typedef struct _GData GData;
...

CS 2506 Computer Organization II C03: Simple MIPS Assembler
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Think carefully about what information would be useful when analyzing and translating the assembly code. Much of this is
actually not part of the source code, but rather part of the specification of the assembly and machine languages. Consider
using static tables of structures to organize language information; by static, I mean a table that's directly initialized when it's
declared, has static storage duration, and is private to the file in which it’s created. For example:

// GData.c
#define NUMRECORDS 50

static GData GISTable[NUMRECORDS] = {
{"New York", "NY", ..., CITY, 33},
{"Pikes Peak", "CO", ..., MOUNTAIN, 14115},
...
{"McBryde Hall", "VA", ..., BUILDING, 2080}
};

Obviously, the sample code shown above does not play a role in my solution. On the other hand, I used this approach to
organize quite a bit of information about instruction formats and encodings. It's useful to consider the difference between
the inherent attributes of an instruction, like its opcode, and situational attributes that apply to a particular occurrence of an
instruction, like the particular registers it uses. Inherent attributes are good things to keep track of in a table. Situational
attributes must be dealt with on a case-by-case basis.

Also, be careful about making assumptions about the instruction formats… Consult the manual MIPS32 Architecture
Volume 2, linked from the Resources page. It has lots of details on machine language and assembly instruction formats. I
found it invaluable, especially in some cases where an instruction doesn't quite fit the simple description of MIPS assembly
conventions in the course notes (e.g., sll and syscall).

Feel free to make reasonable assumptions about limits on things like the number of variables, number of labels, number of
assembly statements, etc. It's not good to guess too low about these things, but making sensible guesses let you avoid
(some) dynamic allocations.

Write lots of "utility" functions because they simplify things tremendously; e.g., string trimmers, mappers, etc.

Data structures play a role because there's a substantial amount of information that must be collected, represented and
organized. However, I used nothing fancier than arrays.

Data types, like the structure shown above, play a major role in a good solution. I wrote a significant number of them.

Explore string.h carefully. Useful functions include strncpy(), strncmp(), memcpy() and strtok(). There
are lots of useful functions in the C Standard Library, not just in string.h. One key to becoming proficient and
productive in C, as in most programming languages, is to take full advantage of the library that comes with that language.

When testing, you should create some small input files. That makes it easy to isolate the various things your assembler
must deal with. Note that the assembler is not doing any validation of the logic of the assembly code, so you don’t have to
worry about producing assembly test code that will actually do anything sensible. For example, you might use a short
sequence of R-type instructions:

.text
add $t0, $t1, $t2
sub $t3, $t1, $t0
xor $s7, $t4, $v0





CS 2506 Computer Organization II C03: Simple MIPS Assembler
Version 8.01 This is a purely individual assignment! 18
Change Log Relative to Version 8.00

If changes are made to the specification, details will be noted below, the version number will be updated, and an
announcement will be made on the course Forum board.

Version Posted Pg Change
8.00 Mar 20 Base document.
8.01 Mar 21 4 Corrected inconsistent description of the assembler's command-line interface.
8.02 Mar 31 9 Corrected description of interface of compare tool.







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