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CIT 593 | Assignment: Dynamic Memory
创建日期：2024-05-17 15:40:03
标签： CIT 593
Dynamic Memory

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CIT 593 | Assignment: Dynamic Memory & File I/O | 1

Setting up Codio for this HW:
1) Open the Codio assignment via Coursera
2) From the Codio File-Tree click on: lc4_memory.h and lc4_memory.c
3) Remember to use the Codio “Pre-submission tests” function before submitting the
assignment.
Overview:
The goal of this HW is for you to write a program that can open and read in a .OBJ file created
by PennSim, parse it, and load it into a linked list that will represent the LC4’s program and data
memories (similar to what PennSim’s “loader” does). In the last HW, you created a .OBJ file. In
this HW, you will be able to read in a .OBJ file and convert it back to the assembly it came from!
This is known as reverse assembling (sometimes a disassembler).
RECALL: OBJECT FILE FORMAT
The following is the format for the binary .OBJ files created by PennSim from your .ASM files. It
represents the contents of memory (both program and data) for your assembled LC-4 Assembly
programs. In a .OBJ file, there are 3 basic sections indicated by 3 header “types” = CODE, DATA,
SYMBOL.
● Code: 3-word header (xCADE,, ), n-word body comprising the instructions.
This corresponds to the .CODE directive in assembly.
● Data: 3-word header (xDADA,, ), n-word body comprising the initial data
values. This corresponds to the .DATA directive in assembly.
● Symbol: 3-word header (xC3B7,, ), n-character body comprising the
symbol string. Note, each character in the file is 1 byte, not 2. There is no null
terminator. Each symbol is its own section. These are generated when you create labels
(such as “END”) in assembly.
LINKED LIST NODE STRUCTURE:
In the file: lc4_memory.h, you’ll see the following structure defined:
The structure is meant to model a row of the LC4’s memory: a 16-bit address, & its 16-bit
contents. As you know, an address may also have a label associated with it. You will also recall
that PennSim always shows the contents of memory in its “assembly” form. So PennSim
reverse-assembles the contents and displays the assembly instruction itself (instead of the
binary contents).
CIT 593 | Assignment: Dynamic Memory & File I/O | 2
As part of this assignment, you will read in a .OBJ file and store each instruction in a NODE of
the type above. Since they’ll be an unknown # of instructions in the file, you’ll create a linked
list of the nodes above to hold all the instructions that are in the .OBJ file.
The details of how to implement all of this will be discussed in the sections of this document
that follow.
CIT 593 | Assignment: Dynamic Memory & File I/O | 3
FLOW CHART: Overview of Program Operation
CIT 593 | Assignment: Dynamic Memory & File I/O | 4
IMPLEMENTATION DETAILS:
The first files to view in the helper file are lc4_memory.h and lc4_memory.c. In these files you
will notice the structure that represents a row_of_memory as referenced above (see the
section: LINKED_LIST_NODE_STRUCTURE above for the node’s layout). You will also see several
helper functions that will serve to manage a linked list of “rows_of_memory” nodes. Your job
will be to implement these simple linked list helper functions using your knowledge from the
last HW assignment. You must implement everything listed by the comments in the starter
code.
Next, you will modify the file called: lc4.c It serves as the “main” for the entire program. The
head of the linked list must be stored in main(), you will see in the provided lc4.c file a pointer
named: memory will do just that. Main() will then extract the name of the .OBJ file the user has
passed in when they ran your program from the argv[] parameter passed in from the user.
Upon parsing that, it will call lc4_loader.c’s open_file() and hold a pointer to the open file. It will
then call lc4_loader.c’s parse_file() to interpret the .OBJ file the user wishes to have your
program process. Lastly it will reverse assemble the file, print the linked list, and finally delete it
when the program ends. These functions are described in greater detail below. The order of
the function calls and their purpose is shown in comments in the lc4.c file that you will
implement as part of this assignment.
Once you have properly implemented lc4.c and have it accept input from the command line, a
user should be able to run your program as follows:
./lc4 my_file.obj
Where “my_file.obj” can be replaced with any file name the user desires as long as it is a valid
.OBJ file that was created by PennSim. If no file is passed in, your program should generate an
error telling the user what went wrong, like this:
error1: usage: ./lc4
There is no need to check that the filename ends in .obj nor should you append .obj to filename
passed in without an extension.
CIT 593 | Assignment: Dynamic Memory & File I/O | 5
Problem 1) Implementing the LC4 Loader
Most of the work of your program will take place in the file: called: lc4_loader.c. In this file, you
will start by implementing the function: open_file() to take in the name of the file the user of
your program has specified on the command line (see lc4_loader.h for the definition of
open_file()). If the file exists, the function should return a handle to that open file, otherwise a
NULL should be returned.
Also in lc4_loader.c, you will implement a second function: parse_file() that will read in and
parse the contents of the open .OBJ file as well as populate the linked_list as it reads the .OBJ
file. The format of the .OBJ input file has been in lecture, but its layout has been reprinted
above (see section: OBJECT_FILE_FORMAT). As shown in the flowchart above, have the function
read in the 3-word header from the file. You’ll notice that all of the LC4 .OBJ file headers consist
of 3 fields: header type,, . As you read in the first header in the file, store the
address field and the field into local variables. Then determine the type of header you have
read in: CODE/DATA/SYMBOL.
If you have read in a CODE header in the .OBJ file, from the file format for a .OBJ file, you’ll
recall the body of the CODE section is -words long. As an example, see the hex listed below,
this is a sample CODE section, notice the field we should correlate with n=0x000C, or decimal:
12. This indicates that the next 12-words in the .OBJ file are in fact 12 LC-4 instructions. Recall
each instruction in LC4 is 1 word long.
CA DE 00 00 00 0C 90 00 D1 40 92 00 94 0A 25 00 0C 0C 66 00 48
01 72 00 10 21 14 BF 0F F8
From the example above, we see that the first LC-4 instruction in the 12-word body is: 9000.
(that happens to be a CONST assembly instruction if you convert to binary). Allocate memory
for a new node in your linked list to correspond to the first instruction (the section above:
LINKED LIST NODE STRUCTURE, declares a structure that will serve as a blue-print for all your
linked list nodes called: “row_of_memory”). As it is the first instruction in the body, and the
address has been listed as 0000, you would populate the row_of_memory structure as follows.
address 0000
label NULL
contents 9000
assembly NULL
next NULL
CIT 593 | Assignment: Dynamic Memory & File I/O | 6
In a loop, read in the remaining instructions from the .OBJ file; allocate memory for a
corresponding row_of_memory node for each instruction. As you create each
row_of_memory add these nodes to your linked list, ordering the list by address (you should
use the functions you’ve created in lc4_memory.c to help you with this). For the first 3
instructions listed in the sample above, your linked list would look like this:
The procedure for reading in the DATA sections would be identical to reading in the CODE
sections. These would become part of the same linked list, as we remember PROGRAM and
DATA are all in one “memory” on the LC-4, they just have different addresses.
For the following SYMBOL header/body:
C3 B7 00 00 00 04 49 4E 49 54
The address field is: 0x0000. The symbol field itself is: 0x0004 bytes long. The next 4 bytes:49
4E 49 54 are ASCII for: INIT. This means that the label for address: 0000 is INIT. Your
program must search the linked list: memory, find the appropriate address that this label is
referring to and populate the “label” field for the node. Note: the field: tells us exactly how
much memory to malloc() to hold the string, however you must add a byte to hold the NULL. 5
bytes in the case of: INIT. For the example above, the node: 0000 in your linked list, would be
updated as follows:
address 0000
label INIT
contents 9000
assembly NULL
next
It might be that you come across a label for an address that has not yet been created in your
linked list. In this case create a new node and add the label. The other node fields can be left
blank. They will be eventually updated. It is possible that an address has two labels in the .OBJ
file. In this case make sure to use the last one that occurs.
CIT 593 | Assignment: Dynamic Memory & File I/O | 7
Once you have read the entire file; created and added the corresponding nodes to your linked
list by address order, close the file and return to main(). If you encounter an error in closing the
file, before exiting, print an error, but also free() all the memory associated with the linked list
prior to exiting the program.
The feof() function may not work the way you think it should. The eof indicator only returns
true AFTER you have attempted to read past the end of the file. If the file position is right at the
end of the file but you haven’t tried to read past the end of the file yet, feof() will be false. This
means that you need to check all of your calls to fread() or fgetc() to make sure they didn’t hit
the end of the file.
CIT 593 | Assignment: Dynamic Memory & File I/O | 8
Problem 2) Implementing the Reverse Assembler
In a new file: lc4_disassembler.c: write a third function (reverse_assemble) that will take as
input the populated “memory” linked list (that parse_file() populated) – it will now contain the
.OBJ’s contents. reverse_assemble() must translate the hex representation of some of the
instructions in the LC4 memory’s linked list into their assembly equivalent. You will need to
reference the LC4’s ISA to author this function. To simplify this problem a little, you DO NOT
need to translate every single HEX instruction into its assembly equivalent. Only translate
instructions with the OPCODE: 0001 (ADD REG, MUL, SUB, DIV, ADD IMM). The immediate value
can be formatted with #, x, X, and/or nothing (e.g. ADD R1, R1, #10 == ADD R1, R1, xF == ADD
R1, R1, 10).
Data stored at an address in the DATA section should not be translated to an assembly
instruction
As shown in the flowchart, this function will call your linked list’s “search_opcode()” helper
function. Your search_opcode() function should take as input a 4-bit opcode and return the first
node in the linked list that matches the opcode passed in, but also has a NULL assembly field.
The opcode that is passed into search_opcode() is a number between 0 and 15 (0x0 and 0xf).
The opcode in the instruction is in the leftmost bits of the instruction. For example, here’s the
definition of an ADD instruction from the ISA:
0001 ddds ss00 0ttt
but the opcode in the opcode parameter passed to search_opcode() will be:
0000 0000 0000 0001
when looking for an instruction with opcode 1. So you will need to use the C-bitwise operators
to line these two values up before comparing them.
search_opcode() is meant to find the first instruction in the linked list where these two 4-bit
fields match. There is an additional constraint that memory rows that already have the
assembly instruction (reverse assembled) filled in don't count as a match.
Said another way, find the first instruction in memory with a matching opcode that hasn't
already been reverse assembled.
When/if a node in your linked list is returned, you’ll need to examine the “contents” field of the
node and translate the instruction into its assembly equivalent. Once you have translated the
contents filed into its ASCII Assembly equivalent, allocate memory for and store this as string in
the “assembly’ field of the node. Repeat this process until all the nodes in the linked list with an
OPCODE=0001 have their assembly fields properly translated.
As an example, the figure below shows a node on your list that has been “found” and returned
when the search_opcode() function was called. From the contents field, we can see that the
HEX code: 128B is 0001 001 010 001 011 in binary. From the ISA, we realize the sub-opcode
reveals that this is actually a MULTIPLY instruction. We can then generate the string MUL R1,
CIT 593 | Assignment: Dynamic Memory & File I/O | 9
R2, R3 and store it back in the node in the assembly field. For this work, I strongly encourage
you to investigate the switch() statement in C (any good book on C will help you understand
how this works and why it is more practical than multiple if/else/else/else statements). I also
remind you that you must allocate memory strings before calling strcpy()!
CIT 593 | Assignment: Dynamic Memory & File I/O | 10
You will find that using C bitwise operators will greatly simplify your implementation. See
https://www.tutorialspoint.com/cprogramming/c_bitwise_operators.htm for details.
Problem 3) Putting it all together
As you may have realized, main() should do only 3 things: 1) create and hold the pointer to your
memory linked list. 2) Call the parsing function in lc4_loader.c. 3) Call the disassembling
function in lc4_dissassembler.c. One last thing to do in main() is to call a function to print the
contents of your linked list to the screen. Call the print_list() function In lc4_memory.c; you will
need to implement the printing helper function to display the contents of your lc4’s memory list
like this:

INIT 0000 9000
0001 D140
0002 9200
…
0009 128B MUL R1, R2, R3
END 000A 0
(and so on…)
Notice thatand fields are displayed in hexadecimal with leading zeroes.
Please remove all debugging print statements before submitting. Only the memory list should
be printed. If the contents of a node are NULL, they should display as 0.
1. You can print (null) for label and assembly instructions when none exist, or leave them
blank.
2. There may be very long labels, so that the other entries in that row are pushed to the
right. That is fine.
3. You may find the formatting strings “%-20.04x” and “%-20s” useful in keeping your
output aligned. This isn’t critical but it is definitely easier to read.
Several things to note: There can be multiple CODE/DATA/SYMBOL sections in one .OBJ file. If
there is more than one CODE section in a file, there is no guarantee that they are in order in
terms of the address. In the file shown above, the CODE section starting at address 0000, came
before the CODE section starting at address: 0010; there is no guarantee that this will always
happen, your code must be able to handle that variation. Also, SYMBOL sections can come
before CODE sections! What all of this means is that before one creates/allocates memory for a
new node in the memory list, one should “search” the list to make certain it does not already
exist. If it exists, update it, if not, create it and add it to the list!
CIT 593 | Assignment: Dynamic Memory & File I/O | 11
Prior to exiting your program, you must properly “free” any memory that you allocated. We will
be using a memory checking program known as valgrind to ensure your code properly releases
all memory allocated on the heap! Simply run your program: lc4 as follows:
valgrind –-leak-check=full ./lc4 test1.obj
where test1.obj is the name of the .obj file you want to test with.
The following section is in bold and red for a reason. Please make sure you fully understand
it!
Valgrind should report 0 errors AND there should be no memory leaks
and no accessing uninitialized memory locations prior to submission.
Note: we will run Valgrind on your submission, if it returns errors, you
will lose points on this assignment. So watch the VIDEO, learn how to
use Valgrind!!
Also note: If your code doesn’t compile or even run, you will lose most
of the points of this assignment!
Valgrind will find errors related to accessing uninitialized memory
locations. This typically results from assuming that malloc() zeroes out
the memory it returns. It does NOT. It can also happen when you
access memory locations in an array or string beyond the end of the
array or string. While your program may run correctly on your test
cases when you have these errors, the results are unpredictable and
may cause the test cases run by the autograder to fail.
TESTING YOUR CODE
When writing such a large program, it is a good strategy to “unit test.” This means, as you
create a small bit of working code, compile it and create a simple test for it. As an example,
once you create your very first function: add_to_list(), write a simple “main()” and test it out.
Call it, print out your “test” list, see if this function even works. Run Valgrind on the code, see if
it leaks memory or accesses uninitialized memory locations. Once you are certain it works, and
doesn’t leak memory, go on to the next function: “search_address()”; implement that, test it
out.
CIT 593 | Assignment: Dynamic Memory & File I/O | 12
DO NOT write the entire program, compile it, and then start testing it. You will never resolve all
of your errors this way. You need to unit test your program as you go along or it will be
impossible to debug.
Remember to use the Codio “Pre-submission tests” function before submitting the
assignment.
Where to get input files?
In the last assignment, you created a .OBJ file. Try loading that file into Codio, and use your
program on it. You know exactly how that program should disassemble. To test further, bring
up PennSim, write a simple program in it, output a .OBJ from PennSim, then read into your
program and see if you can disassemble it. You can create a bunch of test cases very easily with
PennSim.
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CIVL2010 ECMT2150 CS 333 CP1406 KIT308 Economics 701 ECON3740 ECE 5759 BFC3340 FINC2012 COMP4500 MATH3871 FIT3080 18-819C Economic ECO202Y5Y PHY100 FINC-780 Econ 100B EC602 IE582 EE450 COMP9517 WRDS 150B ECON 570 LECTURE 2 ECA5101 ECE 4420 ADAD9311 FIT3031 CS 544 MF 703 CS 6140 KIT501 ENGSCI 344 CSC171 IE 332 FINM8007 21S2 COSC 1114 FE545 E471 SIT102 CSC108 MAT301 IAE 101 COP3503 STAT2004 EC 303 CS1026 STAT1203 MATH3171 ISOM5160 IFN657 FIT2014 ACCT2102 DSCI550 STATS 330 ELEC421 COMP3670 QBUS6840 MEES:698B MAST30027 MIET1077 Economics IFT 6758 FIT5122 BUSN5101 CptS 111 HRMT5502 Math2018 PP 312 INFS3200 CZ3005 CSSE1001 SENG 2011 MKTG 553Q BUFN 640 COMP507 ETF5952 DATA 311 ECE4043 BANK3011 finance M3H623544 HW1 INFS 7410 AI6103 STAT3888 INFS3208 ACCT7103 ECOS3013 PSYCH 20B F79MA STAT3006 ENGG952 2X MGSC 7000 ECE4132 MSF 526 MATH 3033 CDS 511 EEEE4120 EEET2465 ECA5103 MATH2070 ELEC3104 BU51019 EEE8147 ENGR20003 COSC 1107 ETF3200 FNCE 435 MSFI 435 ISyE6420 GENG4405 HW ETC5523 IE 7315 MOEC0021 STA 106 ENGG 1300A ENGG 1300 OLET1139 STAT 3503 STAT 206 FEM11149 ELEN 4720 COMP20003 SOLA1070 CS574 C11CS LINB29H3 MAST90083 ECON7200 STAT 610 LINB29 DSM-5 COMP3161 EEEN40010 MATH 4431 K353 STA 3386 IERG4300 COMS W4167 ENG5298 AMME 2302 CS3910 COMPSCI 320 STATS 369 GR5242 COSC 328 CSCI435 MGTS1301 INFS2200 COMP3900 CIS 375 Module AE03 COMP3121 CS260 HAM503 MGMT5050 INT3075 COSC 285 GEOL0029 GEGE2X01 STAT1201 MATH1061 MMF1941H STAT 5060 CS 520 COMP 4320 MSIN0104 000T FINS3648 ECMT2160 ECON4003 Phys 129L CSE 142 CS 3130 MGEB01 EECS 340 Finance (20443/20444) ECMM149 COMP5310 EEEN60180 CSC 111 AT2 CSCI 4430 STAB57 MA/CSC 427 MATH322 D100 CSCI544 CITS3004 CI6227 CSE 565 APCO 1P01 852L1 ELEC S411F COMP3207 STATS330 ECON7530 FIT5149 CSI4104 CMPT 371 COMP201 SEHS4696 FI4003 MATH6182 CG1 WS MATH401 MGT 205 QTM 110 COMP 0137 MACE12201 FIT5210 ECE 544 BISM7206 HUDM 6055 ECON3203 STATS 210 MSCI 570 MATH6173 COMPSCI4039 Accounting COM3503 AOE2024 ECON213 PSYC10004 Control Data 100/200A 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