CS 2506 Computer Organization II C03: Simple MIPS Assembler 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 , <
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
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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) sw rt, offset(rs) lui rt, imm16
# Transfers a word from memory to a register
# GPR[rt] <-- Mem[GPR[rs] + offset]
# Transfers a word from a register to memory
# Mem[GPR[rs] + offset] <-- GPR[rt]
# 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 addi rt, rs, imm16
addu rd, rs, rt
addiu rt, rs, imm16
and rd, rs, rt andi rt, rs, imm16 mul rd, rs, rt
nop
nor rd, rs, rt sll rd, rt, sa slt rd, rs, rt slti rt, rs, imm16 sra rd, rt, sa Version 8.01
# signed addition of integers; overflow detection
# GPR[rd] <-- GPR[rs] + GPR[rt]
# signed addition with 16-bit immediate;
# overflow detection
# GPR[rt] <-- GPR[rs] + imm16
# unsigned addition with 16-bit immediate;
# no overflow detection
# GPR[rt] <-- GPR[rs] + GPR[rt]
# unsigned addition with 16-bit immediate;
# no overflow detection
# GPR[rt] <-- GPR[rs] + imm16
# bitwise logical AND
# GPR[rd] <-- GPR[rs] AND GPR[rt]
# bitwise logical AND with 16-bit immediate
# GPR[rd] <-- GPR[rs] AND imm16
# signed multiplication of integers;
# no overflow detection
# GPR[rd] <-- (GPR[rs] * GPR[rt])(31:0)
# no operation
# executed as: sll $zero, $zero, 0
# bitwise logical NOR
# GPR[rd] <-- !(GPR[rs] OR GPR[rt])
# logical shift left a fixed number of bits # GPR[rd] <-- GPR[rs] <
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C03: Simple MIPS Assembler
srav rd, rt, rs sub rd, rs, rt
# arithmetic shift right a variable number of bits # GPR[rd] <-- GPR[rt] >>a GPR[rs]
# 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
blez rs, offset
bgtz rs, offset
bne rs, rt, offset
j target syscall
# # #
# # #
# # #
# # #
# #
conditional branch if rs == rt
PC <-- (GPR[rs] == GPR[rt] ? PC + 4 + offset <
PC <-- (GPR[rs] > 0 ? PC + 4 + offset <
CS 2506 Computer Organization II C03: Simple MIPS Assembler
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
CS 2506 Computer Organization II C03: Simple MIPS Assembler 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 num_values: .word sum: .word
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 # array of 10 words
.text
main:
loop:
chk:
la $s0, values
la $s1, num_values lw $s1, 0($s1)
sll $s1, add $s1, li $s2,
j chk
$s1, 2
$s0, $s1
0
lw $s3, 0($s0) add $s2, $s2, $s3
addi $s0, $s0, sub $t0, $s0,
4 $s1
10 0
# size of array
# running total
# load address of the array;
# $s0 will point to the current element
# load address of num_values
# load num_values
# compute size of array in bytes
# compute one-past-end address
# set running total to 0
# check for empty array
# fetch current data element
# update running total
# step pointer to next array element
# compute distance beyond one-past-end
# if ( distance <= 0 ) then loop
blez $t0, loop
# done processing array elements; store total to memory
la $s4, sum
sw $s2, 0($s4)
# report results to user
li $v0, 4
la $a0, message syscall
li $v0, 1
and $a0, $s2, $s2 syscall
li $v0, 10 syscall
# get address for sum
# write sum to memory
# system call to print string
# load address of message into arg register
# make call
# system call to print an integer
# load value to print into arg register
# make call
# system call to terminate program
# make call
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Sample Assembler Output
Here’s my assembler’s output when invoked as: assemble
C03: Simple MIPS Assembler
adder.asm
adder.o:
00100000000100000010000000101100 00100000000100010010000001010100 10001110001100010000000000000000 00000000000100011000100010000000 00000010000100011000100000100000 00100100000100100000000000000000 00001000000000000000000000001010 10001110000100110000000000000000 00000010010100111001000000100000 00100010000100000000000000000100 00000010000100010100000000100010 00011001000000001111111111111011 00100000000101000010000001011000 10101110100100100000000000000000 00100100000000100000000000000100 00100000000001000010000000000000 00000000000000000000000000001100 00100100000000100000000000000001 00000010010100100010000000100100 00000000000000000000000000001100 00100100000000100000000000001010 00000000000000000000000000001100
01010100011010000110010100100000 01110011011101010110110100100000 01101111011001100010000001110100 01101000011001010010000001101110 01110101011011010110001001100101 01110010011100110010000001101001 01101110001000000111010001101000 01100101001000000110000101110010 01110010011000010111100100100000 01101001011100110011101000100000 00000000000000000000000000000000 00000000000000000000000000000010 00000000000000000000000000000011 00000000000000000000000000000101 00000000000000000000000000000111 00000000000000000000000000001011 00000000000000000000000000001101 00000000000000000000000000010001 00000000000000000000000000010011 00000000000000000000000000010111 00000000000000000000000000011101 00000000000000000000000000001010 00000000000000000000000000000000
la
la
lw
sll
add
li
j
lw
add
addi
sub
blez
la
sw
li
la
syscall
li
and
syscall
li
syscall
message +
terminator +
padding
values
num_values
sum
text segment for adder.o
data segment for adder.o
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CS 2506 Computer Organization II C03: Simple MIPS Assembler 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:
00000000000000000000000000000001
00000000000000000000000000000100
S1: 'a' 'b' 'c' '\0'
I1: 0x00000001
S2:
I2: 0x00000004
01100001
01100010
01100011
00000000
01000011
01000100
01000101
01000110
01000111
00000000
0000000000000000
'C'
'G'
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
0x00002000 message
0x0000202C values
0x00002054 num_values
0x00002058 sum
0x00000000 main
0x0000001C loop
0x00000028 chk
-symbols
adder.asm
adder.sym:
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.
'D'
'E'
'F'
'\0'
padding bytes
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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 ./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.
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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==
==27447==
==27447==
==27447==
==27447== Parent PID: 3523
==27447==
Memcheck, a memory error detector
Copyright (C) 2002-2013, and GNU GPL'd, by Julian Seward et al. Using Valgrind-3.10.0 and LibVEX; rerun with -h for copyright info Command: disassem C3TestFiles/ref07.o stu_ref07.asm
==27447==
==27447== HEAP SUMMARY:
==27447==
==27447==
==27447==
==27447==
==27447==
==27447==
==27447==
==27447==
in use at exit: 0 bytes in 0 blocks
total heap usage: 226 allocs, 226 frees, 6,720 bytes allocated
All heap blocks were freed -- no leaks are possible
ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 2 from 2) 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==
==7962==
==7962==
==7962==
==7962== Checked 111,584 bytes
==7962==
==7962==
==7962==
==7962==
==7962==
==7962==
==7962==
==7962==
definitely
indirectly
possibly
lost: 3,116 bytes in 233 blocks
lost: 590 bytes in 96 blocks
lost: 0 bytes in 0 blocks
in use at exit: 4,842 bytes in 331 blocks
total heap usage: 331 allocs, 0 frees, 4,842 bytes allocated
Searching for pointers to 331 not-freed blocks
==7962==
==7962==
==7962==
==7962==
...
==7962== LEAK SUMMARY:
7 bytes in 1 blocks are definitely lost in loss record 1 of 26
at 0x4C2B974: calloc (in /usr/lib64/valgrind/vgpreload_memcheck-amd64-linux.so) by 0x4020B3: parseJTypeInstruction (ParseInstructions.c:178)
by 0x400F51: main (Disassembler.c:137)
still reachable: 1,136 bytes in 2 blocks suppressed: 0 bytes in 0 blocks
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.
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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 addi rt, rs, imm16
nor rd, rs, rt slti rt, rs, imm16 syscall
# signed addition of integers; overflow detection
# GPR[rd] <-- GPR[rs] + GPR[rt]
# signed addition with 16-bit immediate;
# overflow detection
# GPR[rt] <-- GPR[rs] + imm16
# bitwise logical NOR
# GPR[rd] <-- !(GPR[rs] OR GPR[rt])
# set register to result of comparison
# GPR[rt] <-- (GPR[rs] < imm16 ? 0 : 1)
# 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) la rd, label
# Transfers a word from memory to a register.
# GPR[rt] <-- Mem[GPR[rs] + offset]
# load address label to register
# GPR[rd] <-- label
# pseudo-translation for 16-bit label:
# addi rd, $zero, label
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mul rd, rs, rt # signed multiplication of integers; # no overflow detection
nor rd, rs, rt
beq rs, rt, offset
bne rs, rt, offset
# GPR[rd] <-- (GPR[rs] * GPR[rt])(31:0)
# bitwise logical NOR
# GPR[rd] <-- !(GPR[rs] OR GPR[rt])
# conditional branch if rs == rt
# PC <-- (GPR[rs] == GPR[rt] ? PC + 4 + offset <
C3TestFiles/
C3TestFiles/mref01.asm
C3TestFiles/mref01.o
C3TestFiles/mref02.asm
C3TestFiles/mref02.o
C3TestFiles/mref03.asm
C3TestFiles/mref03.o
compare
prepC3.sh
testC3.sh
Some General Coding Requirements
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
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).
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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.
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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.
//
//
//
//
Credits
Failure to include the pledge statement may result in your submission being ignored.
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.
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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; …
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CS 2506 Computer Organization II C03: Simple MIPS Assembler
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
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CS 2506 Computer Organization II C03: Simple MIPS Assembler
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
8.00 Mar 20
8.01 Mar 21
Pg Change
Base document.
4 Corrected inconsistent description of the assembler’s command-line interface.
Version 8.01
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