代写代考 CIT593| Assignment:C-FileI/O– Making the LC4Assembler

CIT593| Assignment:C-FileI/O– Making the LC4Assembler
Setting up Codio for this HW:
1) Open the Codio assignment via Coursera
2) From the Codio File-Tree click on: assembler.c

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From lecture you’ve learned that C is file-oriented and that working with files represents I/O devices in C. C places files into two categories: “text” and “binary.” In this assignment you’ll work with both types by reading in a text file and writing out a binary file. The text file that you will read in this assignment will be a .ASM file (a text file intended for PennSim) and the type of output file you’ll generate will be a .OBJ file ( the same type of binary file that PennSim would write out). Aside from reading and writing out the files, your task will be make a mini-LC4- Assembler! A program that reads in assembly language and generates its machine equivalent. This assignment will require a bit more programming rigor than we’ve had thus far, but now that you’ve gained a good amount of programming skill in this class and in others, it is the perfect time to tackle a large programming assignment.
If your program is crashing, before you can be helped via TA office hours, you must run GDB on
your program. A small amount of detective work with GDB will save hours! Be prepared for
TAs to tell you to run GDB before they will help you in office hours. Watch the tutorial and run
it on your program if it is crashing
Assignment Overview:

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Problem #1: Reading in a text file (the .ASM file)
Open “assembler.c” from the helper files; it contains the main() function for the program. Carefully examine the variables at the top:
char* filename = NULL ;
char program [ROWS][COLS] ;
char program_bin_str [ROWS][17] ;
unsigned short int program_bin [ROWS] ;
The first pointer variable “filename” will be a pointer to a string that contains the text file you’ll be reading. Your program must take in as an argument the name of a .ASM file. As an example, once you compile your main() program, you would execute it as follows:
./assembler test.asm
In your last HW you learned how to use the arguments passed into main(). So the first thing to implement is to check if argc has arguments, and if it does, point “filename” to the argument that contains the passed in string that is the file’s name. You should return from main() immediately (with an error message) if the caller doesn’t provide an input file name as follows:
error1: usage: ./assembler
Start by updating “assembler.c” to read in the arguments. Compile your changes and test them before continuing. You should take this moment to setup your Makefile as well, it should contain two basic directives: assembler and asm_parser.o
After you’ve successfully gotten the filename from the caller, the first function you must call will be:
int read_asm_file (char* filename, char program [ROWS][COLS] ) ;
The purpose of read_asm_file() is to open the ASM file, and place its contents into the 2D array: program[][]. You must complete the implementation of this function in the provided helper file: “asm_parser.c”. Notice, it takes in the pointer to the “filename” that you’ll open in this function. It also takes in the two dimensional array, program, that was defined back in main(). You’ll see that “ROWS” and “COLS” are two #define’ed constants in the file: asm_parser.h. Rows is set to 100 and COLS is set to 255. This means that you can only read in a program that is up to 100 lines long and each line of this program can be no longer than 255.
You’ll want to look at the class notes (or a C-reference textbook) to use fopen() to open the filename that has been passed in. Then you’ll want to use a function like: fgets() to read each

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line of the .ASM file into the program[][] 2D array. Be aware that “fgets()” will keep carriage returns (aka – the newline character) and you’ll need to strip these from the input.
Take a look at test.asm file that was included in the helper file. It contains the following program:
ADD R1, R0, R1
MUL R2, R1, R1
SUB R3, R2, R1
DIV R1, R3, R2
AND R1, R2, R3
OR R1, R3, R2
XOR R1, R3, R2
After you complete read_asm_file() and if you were to run it on test.asm, your 2D array: program[][]should contain the contents of the ASM file in this order:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
6XO R 7 ‘\0’ X X
R1 , R2 , R3 , R1 , R1 ,
R0 , R1 , R2 , R3 , R2 ,
R 1 ‘\0’ R1‘\0’ R1‘\0’ R2‘\0’ R3‘\0’ 2 ‘\0’ X R2‘\0’ X X X
Notice, there are no “newline” characters at the end of the lines.
If reading in the file is a success, return 0 from the function, if not, return 2 from the function
and print an error to the screen: error2: read_asm_file() failed Implement and test this function carefully before continuing on with the assignment.

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Problem #2: Parsing an Instruction
Once read_asm_file() is working properly, back in main(), you’ll call the function: parse_instruction(), which is also located in asm_file.c:
int parse_instruction (char* instr, char* instr_bin_str) ;
purpose, arguments & return value
The purpose of this function is to take in a single row of your program[][] array and convert to its binary equivalent – in text form. The argument: instr must point to a row in main()’s 2D array: program[][]. The argument: instr_bin_str must point to the corresponding row in main()’s 2D array: program_bin_str[][]. If there no errors are encountered the function will return a 0 and if any error occurs (in this function) it should return the number 3 and an error message should be printed: error3: parse_instruction() failed.
Let’s assume you’ve called parse_instruction() and “instr” points to the first row in your program[][] array:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 *intr-> A D D R 1 , R 0 , R 1 ‘\0’
Parse_instruction() needs to examine this string and convert it into a binary equivalent. You’ll need to use the LC4 ISA to determine the binary equivalent of an instruction. When your function returns, the memory pointed to by: instr_bin_str, should look like this:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 *intr_bin_str-> 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 ‘\0’
Notice, this isn’t actually binary, but it is the “ADD” instruction’s binary equivalent in TEXT form. We will convert this string form of the binary instruction to HEX later in the function: convert_instrunction().

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How to implement this function
The purpose of converting the instruction to a binary string (instead of to the binary # it will eventually become), is so that you can build this string up little by little. Investigate the “strtok()” function in the standard C string library: https://www.tutorialspoint.com/c_standard_library/c_function_strtok.htm
STRTOK() allows you to parse a string that is separated by “tokens”. In this function you’ll be parsing the string pointed to be “instr” and you’ll be building up the string pointed to by “instr_bin_str”. “instr” will contain spaces and commas (those will be your tokens). Your first call to strtok() on the “instr” string should return back the OPCODE: ADD, SUB, MUL, DIV, XOR, etc. The only thing common to all 26 instructions in the ISA is that the very first part of them is the opcode. Once you determine the OPCODE, you’ll call the appropriate helper function to parse the remainder of the instruction.
As an example, let’s say the OPCODE is ADD. For this problem, you do not need to worry about the “immediate” variants of the ADD instruction (or AND immediate) – when we test your code, we won’t use ADD – immediate or AND immediate instructions. Once you’ve determined the OPCODE is ADD, you would call the parse_add() helper function. It will take the instruction (instr) as an argument, but also the instr_bin_str string because parse_add will be responsible for determining the binary equivalent for the ADD instruction you are currently working on and it will update instr_bin_str.
int parse_add (char* instr, char* instr_bin_str ) ;
When parse_add() returns, if no errors occurred during parsing the ADD instrunction, instr_bin should now be complete. At this time, you can return a 0 from parse_instrunction().

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Problem #3: Parsing an ADD instruction
int parse_add (char* instr, char* instr_bin_str ) ;
The helper function: parse_add(), should be called only by the parse_instruction() function. It has two char* arguments: instr and instr_bin_str. Because this function will only be called when an ADD OPCODE is encountered by parse_instruction(), instr will contain an ADD instruction and instr_bin_str should be empty. [][]. If there no errors are encountered the function will return a 0 and if any error occurs (in this function) it should return the number 4 and an error message should be printed: error4: parse_add() failed.
The purpose of this function is to populate the instr_bin_str. Upon the function’s start, the binary OPCODE can be immediately copied into the instr_bin_str[0:3]. Afterwards, the strtok() function can be used again to separate the registers: RD, RS, RT, from the “instr” string.
For each register: RD, RS, RT, the parse_reg() helper function should be called:
int parse_reg (char reg_num, char* instr_bin_str) ;
This function must take as input a number in character form and populate instr_bin_str with the appropriate corresponding binary number. For example, if RD = R0 for the ADD instruction, the ‘0’ character would be passed in the argument: reg_num. parse_reg() would then copy into instr_bin_str[4:6] the characters ‘000’. A 5 should be returned if any errors occur and the standard error message should be printed; otherwise a 0 should be returned upon success.
To implement the parse_reg() function, you should consider using a switch() statement:
https://www.tutorialspoint.com/cprogramming/switch_statement_in_c.htm
This helper function: parse_reg() should only parse one register at a time. Also, because it is not specific to the ADD instruction (nearly all instructions contain registers), it can be called from other functions that need their register’s converted to binary. Example: prase_mul() should also call parse_reg().
Note that parse_add() must also populate the SUB-OPCODE field in instr_bin_str[10:12]. When parse_add() returns, instr_bin_str should be complete. parse_instrunction() should then return to main(). You will need to create a helper function for each instruction type, use parse_add() as a model. As an example, you’ll need to create: parse_mul(), parse_xor(), etc – they will all be very similar functions, so perfect parse_add() before you attempt those other functions.

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Problem #4: Converting the binary string to hex
After parse_instruction() returns successfully to main, the function:
unsigned short int str_to_bin (char* instr_bin_str) ;
Should be called from main(), passing the recently parsed binary string from the array: program_bin_str[X], where “X” represents the binary instruction that was just populated by the last call to parse_instruction().
The purpose of this function is to take a binary string and convert it to a 16-bit binary equivalent and return it to the calling function. To implement this function, the string function, strtol() is recommended:
https://www.tutorialspoint.com/c_standard_library/c_function_strtol.htm
If a zero is returned from strtol(), return the error #6 should be returned and printed to the screen.
As an example of what this function should do, if it was called with the following argument:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 *intr_bin_str-> 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 ‘\0’
then it should return: 0x1201, that is the hex equivalent for this binary string. You can verify and print out what it returns by using printf (“0x%X”), which will print out integers in HEX format.
Once str_to_bin() is returned, it should be assigned to the corresponding spot in the unsigned short int array: program_bin[X], where X matches the index from program_bin_str[X].

…back in main():
The sequence of events in main() should be as follows:
Initialize all arrays to zero or ‘\0’
Call read_asm_file() to read the entire ASM file into the array: program[][]. Using test.asm as an example, after read_asm_file() returns: program[][] should then contain:
0 1 2 0 AD D 1 MU L 2 SU B 3 DI V 4 AN D
R1 , R2 , R3 , R1 , R1 ,
7 8 9 10 11 12 13 14
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5 OR 6XO R 7 ‘\0’ X X
R0 , R1 , R2 , R3 , R2 ,
R 1 ‘\0’ R1‘\0’ R1‘\0’ R2‘\0’ R3‘\0’ 2‘\0’X R2‘\0’ X X X
In a loop, for each row X in program[][] {
3) Call parse_instruction(), passing it the current row in program[X][] as input to parse_instruction(). When parse_instruction() returns program_bin_str[X][] should be updated to have the binary equivalent (in string form).
4) Call str_to_bin() passing program_bin_str[X][] to it. Upon str_to_bin()’s return, program_bin[X] should be updated to have the hex equivalent of the binary string from program_bin_str[X].
} once the loop is complete program_bin_str[][] should contain program[][] equivalent:
0 12345678910111213141516
0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 ‘\0’
1 0 0 0 1 0 1 0 0 0 1 0 0 1 0 0 1 ‘\0’
2 0 0 0 1 0 1 1 0 1 0 0 1 0 0 0 1 ‘\0’
3 0 0 0 1 0 0 1 0 1 1 0 1 1 0 1 0 ‘\0’
4 0 1 0 1 0 0 1 0 1 0 0 0 0 0 1 1 ‘\0’
5 0 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 ‘\0’
6 0 1 0 1 0 0 1 0 1 1 0 1 1 0 1 0 ‘\0’ 7 ‘\0’ X X X X X X X X X X X X X X X X
Also after the loop is complete, the array: program_bin[] should contain program_bin_str[][]’s equivalent in binary (it’s shown in hex here):
0 0x1201 1 0x1449 2 0x1691 3 0x12DA 4 0x5283 5 0x52D2 6 0x52DA
program_bin[] now represents the completely assembled program and can be written out to a .OBJ file in binary form.

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Problem #5: Writing out the object file
During lecture, we learned that a .OBJ file is a binary file. The format of the .OBJ file is discussed in lecture 11, slides 12-13. Recall from lecture that the .OBJ file’s “CODE” header is as follows:
● Code: 3-word header (xCADE,