Difference between revisions of "CSC231 Homework 2 2017"
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sub dest, source ; dest <-- dest - source | sub dest, source ; dest <-- dest - source | ||
| − | '''Note 2''': Do not use the mul instruction. It is too complicated for us to use now. Instead, if you want to multiply something by two, just add that quantity twice. In other words, if you want to compute 2*x, then simply compute x + x. Similarly, when you need to compute 3* | + | '''Note 2''': Do not use the mul instruction. It is too complicated for us to use now. Instead, if you want to multiply something by two, just add that quantity twice. In other words, if you want to compute 2*x, then simply compute x + x. Similarly, when you need to compute 3*z, simply compute z + z + z. |
You should add code '''only between the two box comments''' shown below: | You should add code '''only between the two box comments''' shown below: | ||
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==Examples== | ==Examples== | ||
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Revision as of 15:07, 15 February 2017
--D. Thiebaut (talk) 16:14, 9 February 2017 (EST)
Contents
This assignment is due on 2/20/17 at 11:55 p.m.
Problem #1
Answer the quizzes in the Moodle Homework 2 section.
Problem #2
In your 231b-xx account on aurora, run the following commands:
getcopy 231Lib.asm getcopy hw2_skel.asm cp hw2_skel.asm hw2.asm
You should now have a new file called hw2.asm in your directory, and another file called 231Lib.asm. 231Lib.asm is a library we will use to print and input decimal numbers. The library file will help us get input from the keyboard, and print strings. You do not need to understand the code in 231Lib.asm in order to use it!
The hw2.asm file is just a skeleton and you will need to add code to it to solve this problem. For right now, just assemble and link it to the new library, and run it:
nasm -f elf hw2.asm nasm -f elf 231Lib.asm ld -melf_i386 -o hw2 hw2.o 231Lib.o
The program will prompt you for 3 integer numbers, which it will put into three 32-bit integers called a, b, and c in the data segment. It then takes the integer stored in a fourth variable called ans (for answer), and prints it. Since ans is initialized with 0, that's the number that gets printed.
Here's an example of what happens if I run the program and feed it 3 numbers: 1, 2, and 3 (user input underlined):
./hw2 > 1 > 2 > 3 ans = 0
Your Assignment
Modify hw2.asm and make it compute ans = 2*(a-b) + 3*c.
Note 1: you will need to use the sub instruction, which works similarly to add:
sub dest, source ; dest <-- dest - source
Note 2: Do not use the mul instruction. It is too complicated for us to use now. Instead, if you want to multiply something by two, just add that quantity twice. In other words, if you want to compute 2*x, then simply compute x + x. Similarly, when you need to compute 3*z, simply compute z + z + z.
You should add code only between the two box comments shown below:
;; ----------------------------------- ;; computation: ans = 2*(a-b) + 3*c ;; ----------------------------------- ; your code will go here... ;; ----------------------------------- ;; display "ans =" ;; -----------------------------------
Examples
I ran my solution program a few times with different numbers. Your program should behave exactly the same!
231b@aurora ~/hw/hw2 $ ./hw2 > 3 > 1 > 10 ans = 34 231b@aurora ~/hw/hw2 $ ./hw2 > 10 > 10 > 0 ans = 0 231b@aurora ~/hw/hw2 $ ./hw2 > 1 > 2 > 3 ans = 7 231b@aurora ~/hw/hw2 $ ./hw2 > 10 > 0 > 10 ans = 50
Submission
Submit your program on Moodle, in the Problem 2 section.
Assembly File
Below are the two files used in Problem 2, for reference.
hw2.asm
;;; hw2.asm ;;; put your name here ;;; describe what the program does ;;; explain how to assemble and run it. extern _printDec extern _printString extern _println extern _getInput section .data prompt db "> " promptLen equ $-prompt ansStr db "ans = " ansStrLen equ $-ansStr a dd 0 b dd 0 c dd 0 ans dd 0 section .text global _start _start: ;; display prompt mov ecx, prompt mov edx, promptLen call _printString ;; get a call _getInput mov dword[a], eax ;; display prompt mov ecx, prompt mov edx, promptLen call _printString ;; get b call _getInput mov dword[b], eax ;; display prompt mov ecx, prompt mov edx, promptLen call _printString ;; get c call _getInput mov dword[c], eax ;; ----------------------------------- ;; computation: ans = 2*(a-b) + 3*c ;; ----------------------------------- ; your code will go here... ;; ----------------------------------- ;; display "ans =" ;; ----------------------------------- mov ecx, ansStr mov edx, ansStrLen call _printString ;; ----------------------------------- ;; display ans variable ;; ----------------------------------- mov eax, dword[ans] call _printDec call _println call _println ;;; exit mov ebx, 0 mov eax, 1 int 0x80
231Lib.asm
;;; 231Lib.asm ;;; A simple I/O library for CSC231. ;;; will be expanded as needed. ;;; ;;; D. Thiebaut ;;; Adapted from Swarnali Ahmed's 2002 program: mytools.inc ;;; http://cs.smith.edu/dftwiki ;;; ;;; Contains several functions for performing simple I/O of ;;; data. ;;; _printDec: function that prints an integer on the screen ;;; _printString: function that prints a string on the screen ;;; _println: moves the cursor to the next line on the screen ;;; _getInput: gets a possibly signed integer from the keyboard ;;; ;;; Version 2. Sept 22, 2014. ;;; - updated _getInput to get only 1 char at at time. ;;; - now works with pipes ;;; Version 1. Sept 21, 2014. ;;; %assign SYS_EXIT 1 %assign SYS_WRITE 4 %assign STDOUT 1 global _printDec global _printString global _println global _getInput global _printRegs global _printHex section .text ;;; ;------------------------------------------------------ ;;; ;------------------------------------------------------ ;;; ; getInput: gets a numerical input from the keyboard. ;;; ; returns the resulting number in eax (32 bits). ;;; ; recognize - as the first character of ;;; ; negative numbers. Does not skip whitespace ;;; ; at the beginning. Stops on first not decimal ;;; ; character encountered. ;;; ; ;;; ; NO REGISTERS MODIFIED, except eax ;;; ; ;;; ; Example of call: ;;; ; ;;; ; call getInput ;;; ; mov dword[x], eax ; put integer in x ;;; ; ;;; ;------------------------------------------------------ ;;; ;------------------------------------------------------ _getInput: section .bss buffer resb 120 intg resd 1 isneg resb 1 section .text pushad ; save all registers mov esi, buffer ; eci --> buffer mov edi, 0 ; edi = counter of chars .loop1: mov eax, 03 ; input mov ebx, 0 ; stdin mov ecx, esi ; where to put the next char mov edx, 1 ; one char at a time int 0x80 ; get the input into buffer cmp byte[esi], 0 ; EOF? je .parse cmp byte[esi], 10 ; line feed? je .parse inc esi ; point to next cell inc edi ; increment char counter jmp .loop1 .parse: mov esi, buffer ; esi --> buffer mov ecx, edi ; loop for all chars received mov dword[intg], 0 mov byte[isneg], 0 .negativ: cmp byte[esi], '-' jne .loop inc byte[isneg] .loop: mov ebx, 0 mov bl, byte[esi] ;; stop on line feed cmp bl, 10 ; line feed? je .done ;; stop on non-digit characters cmp bl, '0' jb .done cmp bl, '9' ja .done ;; bl is a digit... multiply .int by 10 first mov edx, 10 mov eax, dword[intg] mul edx ; edx:eax <-- 10 * .int ;; add int version of char sub bl, '0' add eax, ebx mov dword[intg], eax inc esi loop .loop .done: ;; if negative, make eax neg cmp byte[isneg], 0 je .return neg eax mov dword [intg], eax ;; restore registers and return result in eax .return: popad mov eax, [intg] ret ;;; ;------------------------------------------------------ ;;; ;------------------------------------------------------ ;;; ; _printDec: takes the double word in eax and prints it ;;; ; to STDOUT in decimal. ;;; ; ;;; ; Examples: ;;; ; print a byte variable ;;; ; mov eax, 0 ;;; ; mov al, byte[someVar] ;;; ; call _printDec ;;; ; ;;; ; print a word variable ;;; ; mov eax ;;; ; mov ax, word[otherVar] ;;; ; call _printDec ;;; ; ;;; ; print a double-word variable ;;; ; mov eax, dword[thirdVar] ;;; ; call _printDec ;;; ; ;;; ; print register edx in decimal ;;; ; ;;; ; mov eax, edx ;;; ; call _printDec ;;; ; ;;; ;REGISTERS MODIFIED: NONE ;;; ;------------------------------------------------------ ;;; ;------------------------------------------------------ _printDec: ;;; saves all the registers so that they are not changed by the function section .bss .decstr resb 10 .ct1 resd 1 ; to keep track of the size of the string section .text pushad ; save all registers mov dword[.ct1],0 ; assume initially 0 mov edi,.decstr ; edi points to decstring add edi,9 ; moved to the last element of string xor edx,edx ; clear edx for 64-bit division .whileNotZero: mov ebx,10 ; get ready to divide by 10 div ebx ; divide by 10 add edx,'0' ; converts to ascii char mov byte[edi],dl ; put it in sring dec edi ; mov to next char in string inc dword[.ct1] ; increment char counter xor edx,edx ; clear edx cmp eax,0 ; is remainder of division 0? jne .whileNotZero ; no, keep on looping inc edi ; conversion, finish, bring edi mov ecx, edi ; back to beg of string. make ecx mov edx, [.ct1] ; point to it, and edx gets # chars mov eax, SYS_WRITE ; and print! mov ebx, STDOUT int 0x80 popad ; restore all registers ret ;;; ; ------------------------------------------------------------ ;;; ; _printString: prints a string whose address is in ;;; ; ecx, and whose total number of chars ;;; ; is in edx. ;;; ; Examples: ;;; ; Assume a string labeled msg, containing "Hello World!", ;;; ; and a constant MSGLEN equal to 12. To print this string: ;;; ; ;;; ; mov ecx, msg ;;; ; mov edx, MSGLEN ;;; ; call _printSTring ;;; ; ;;; ; REGISTERS MODIFIED: NONE ;;; ; ------------------------------------------------------------ ;;; ;save eax and ebx so that it is not modified by the function _printString: push eax push ebx mov eax,SYS_WRITE mov ebx,STDOUT int 0x80 pop ebx pop eax ret ;;; ; ------------------------------------------------------------ ;;; ; _println put the cursor on the next line. ;;; ; ;;; ; Example: ;;; ; call _println ;;; ; ;;; ; REGISTERS MODIFIED: NONE ;;; ; ------------------------------------------------------------ _println: section .data .nl db 10 section .text push ecx push edx mov ecx, .nl mov edx, 1 call _printString pop edx pop ecx ret ;;; ; ------------------------------------------------------------ ;;; ; _printHex: prints contents of eax in hexadecimal, uppercase. ;;; ; using 8 chars. ;;; ; ------------------------------------------------------------ section .data table db "0123456789ABCDEF" hexString db "xxxxxxxx" tempEax dd 0 section .text _printHex: pushad ; save all registers mov [tempEax], eax ; save eax, as we are going to need it ; several times, for all its digits mov ecx, 8 ; get ready to loop 8 times ;; get char equivalent to lower nybble of eax for: and eax, 0xF ; isolate lower nybble of eax add eax, table ; and translate it into ascii hex char mov bl, byte[eax] ; bl <- ascii ;; make eax point to place where to put this digit in hexString mov eax, hexString ; beginning of table add eax, ecx ; add ecx to it, since ecx counts dec eax ; but eax 1 too big, decrement it mov byte[eax], bl ; store ascii char at right index in hexString ;; shift eax down by 4 bits by dividing it by 16 mov ebx, 16 ; ready to shift down by 4 bits mov eax, [tempEax] ; get eax back div ebx ; shift down by 4 mov [tempEax], eax ; save again loop for ; and repeat, 8 times! ;; print the pattern in hexString, which contains 8 chars mov ecx, hexString ; mov edx, 8 call _printString popad ret ;;; ; ------------------------------------------------------------ ;;; ; _printInt: prints the contents of eax as a 2's complement ;;; ; number. ;;; ; ------------------------------------------------------------ _printInt: push eax ; save the regs we are using push ecx push edx ;; check if msb is set cmp eax, 0x8000000 jb positive push eax ;; the number is negative. Print a minus sign and ;; the positive equivalent of the number mov ecx, minus mov edx, 1 call _printString pop eax neg eax ;; the number is positive, just print it. positive: call _printDec pop edx pop ecx pop eax ret ;;; ; ------------------------------------------------------------ ;;; ; printMinus, _printSpace: print a minus sign, and a plus sign. ;;; ; ------------------------------------------------------------ _printMinus: push ecx push edx mov ecx, minus mov edx, 1 call _printString pop edx pop ecx ret _printSpace: push ecx push edx mov ecx, space mov edx, 1 call _printString pop edx pop ecx ret ;;; ; ------------------------------------------------------------ ;;; ; _printRegs: prints all the registers. ;;; ; ------------------------------------------------------------ section .data minus db '-' space db ' ' eaxStr db 'eax ' ebxStr db 'ebx ' ecxStr db 'ecx ' edxStr db 'edx ' esiStr db 'esi ' ediStr db 'edi ' eaxTemp dd 0 ebxTemp dd 0 ecxTemp dd 0 edxTemp dd 0 esiTemp dd 0 ediTemp dd 0 section .text _printRegs: mov [eaxTemp], eax mov [ebxTemp], ebx mov [ecxTemp], ecx mov [edxTemp], edx mov [ediTemp], edi mov [esiTemp], esi pushad ;; print eax mov ecx, eaxStr mov edx, 4 call _printString call _printHex call _printSpace call _printDec call _printSpace call _printInt call _println ;; print ebx mov ecx, ebxStr mov edx, 4 call _printString mov eax, [ebxTemp] call _printHex call _printSpace call _printDec call _printSpace call _printInt call _println ;; print ecx mov ecx, ecxStr mov edx, 4 call _printString mov eax, [ecxTemp] call _printHex call _printSpace call _printDec call _printSpace call _printInt call _println ;; print edx mov ecx, edxStr mov edx, 4 call _printString mov eax, [edxTemp] call _printHex call _printSpace call _printDec call _printSpace call _printInt call _println ;; print edi mov ecx, ediStr mov edx, 4 call _printString mov eax, [ediTemp] call _printHex call _printSpace call _printDec call _printSpace call _printInt call _println ;; print esi mov ecx, esiStr mov edx, 4 call _printString mov eax, [esiTemp] call _printHex call _printSpace call _printDec call _printSpace call _printInt call _println popad ret
