Difference between revisions of "CSC231 Homework Solutions 2017"
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::<source lang="asm"> | ::<source lang="asm"> | ||
| − | ;;; ; | + | ;;; ; hw5_1.asm |
| − | ;;; ; | + | ;;; ; D. Thiebaut |
;;; ; | ;;; ; | ||
| − | ;;; ; | + | ;;; ; Prints 10 lines of Pascal triangle. |
;;; ; | ;;; ; | ||
;;; ; to assemble and run: | ;;; ; to assemble and run: | ||
;;; ; | ;;; ; | ||
| − | ;;; ; nasm -f elf -F stabs | + | ;;; ; nasm -f elf -F stabs hw5_1.asm |
| − | ;;; ; ld -melf_i386 -o | + | ;;; ; ld -melf_i386 -o hw5_1 hw5_1.o |
| − | ;;; ; ./ | + | ;;; ; ./hw5_1 |
;;; ; ------------------------------------------------------------------- | ;;; ; ------------------------------------------------------------------- | ||
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</source> | </source> | ||
<br /> | <br /> | ||
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=hw5_2.sh= | =hw5_2.sh= | ||
<br /> | <br /> | ||
Revision as of 17:26, 22 April 2017
--D. Thiebaut (talk) 17:57, 22 April 2017 (EDT)
Problem 1
;;; mystery.asm ;;; D. Thiebaut ;;; ;;; To assemble, link, and run: ;;; nasm -f elf mystery.asm ;;; ld -melf_i386 -o mystery mystery.o ;;; ./mystery ;;; section .data Hello db "Hello there!", 10, 10 HelloLen equ $-Hello section .text global _start _start: ;;; print Hello and a space after it mov eax, 4 ; write mov ebx, 1 ; stdout mov ecx, Hello ; address of message to print mov edx, 6 ; print only 6 chars int 0x80 ;;; print ! and two line feed chars mov eax, 4 ; write mov ebx, 1 ; stdout mov ecx, Hello+11 ; address of ! char mov edx, 3 ; print only 3 chars int 0x80 ;;; exit mov ebx, 0 mov eax, 1 int 0x80
hw1.asm
;;; hw1.asm ;;; D. Thiebaut ;;; ;;; To assemble, link, and run: ;;; nasm -f elf hw1.asm ;;; ld -melf_i386 -o hw1 hw1.o ;;; ./hw1 ;;; section .data msg db " CSC231 Assembly " linefeed db 10 dashline db "------------------" lineLen equ $-dashline section .text global _start _start: ;;; print message mov eax, 4 ; write mov ebx, 1 ; stdout mov ecx, dashline ; address of message to print mov edx, lineLen ; # of chars to print int 0x80 mov eax, 4 ; write mov ebx, 1 ; stdout mov ecx, dashline ; address of message to print mov edx, lineLen ; # of chars to print int 0x80 mov eax, 4 ;print line-feed mov ebx, 1 mov ecx, linefeed mov edx, 1 int 0x80 mov eax, 4 ; write ---- mov ebx, 1 ; stdout mov ecx, dashline ; address of message to print mov edx, 9 ; # of chars to print int 0x80 mov eax, 4 ; write message mov ebx, 1 ; stdout mov ecx, msg ; address of message to print mov edx, lineLen ; # of chars to print int 0x80 mov eax, 4 ; write ---- mov ebx, 1 ; stdout mov ecx, dashline ; address of message to print mov edx, 9 ; # of chars to print int 0x80 mov eax, 4 ;print line-feed mov ebx, 1 mov ecx, linefeed mov edx, 1 int 0x80 mov eax, 4 ; write mov ebx, 1 ; stdout mov ecx, dashline ; address of message to print mov edx, lineLen ; # of chars to print int 0x80 mov eax, 4 ; write mov ebx, 1 ; stdout mov ecx, dashline ; address of message to print mov edx, lineLen ; # of chars to print int 0x80 mov eax, 4 ;print line-feed mov ebx, 1 mov ecx, linefeed mov edx, 1 int 0x80 ;;; exit mov ebx, 0 mov eax, 1 int 0x80
hw2.asm
;;; ; ; hw2sol.asm ;;; ; ; D. Thiebaut ;;; ; ; ;;; ; ; 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
Teller.asm
;;; ; teller.asm ;;; ; D. Thiebaut ;;; ; ;;; ; Demo of a teller machine program. ;;; ; To assemble, link, and run: ;;; ; nasm -f elf teller.asm ;;; ; ld -melf_i386 -o teller teller.o 231Lib.o ;;; ; ./teller ;;; ; ;;; extern functions that will be linked to this program ;;; ; contained in 231Lib.asm extern _printDec extern _printString extern _println extern _getInput ;;; ------------------------------------------------------ ;;; DATA SECTION ;;; ------------------------------------------------------ section .data amount dd 139 no20s dd 0 no10s dd 0 no5s dd 0 no1s dd 0 msg db "amount? " MSGLEN equ $-msg ;;; ------------------------------------------------------ ;;; CODE SECTION ;;; ------------------------------------------------------ section .text global _start _start: ;;; get amount from the user mov ecx, msg mov edx, MSGLEN call _printString call _getInput mov dword[amount], eax call _println ;;; Break down amount in 20s, 10s, 5s, and 1s mov eax, dword[amount] mov edx, 0 mov ebx, 20 mov eax, dword[amount] call break mov dword[no20s], eax mov ebx, 10 mov eax, dword[amount] call break mov dword[no10s], eax mov ebx, 5 mov eax, dword[amount] call break mov dword[no5s], eax mov eax, dword[amount] mov dword[no1s], eax mov eax, dword[no20s] call _printDec call _println mov eax, dword[no10s] call _printDec call _println mov eax, dword[no5s] call _printDec call _println mov eax, dword[no1s] call _printDec call _println ;;; exit mov ebx, 0 mov eax, 1 int 0x80 ;;; sum: gets 2 ints in eax and ebx and returns the sum in ;;; eax. sum: add eax, ebx, ret ;;; break: gets amount in eax, divider in ebx, divides, ;;; and puts remainder in [amount] and quotient in eax break: mov edx, 0 div ebx mov dword[amount], edx ret
hw5_1.asm
;;; ; hw5_1.asm ;;; ; D. Thiebaut ;;; ; ;;; ; Prints 10 lines of Pascal triangle. ;;; ; ;;; ; to assemble and run: ;;; ; ;;; ; nasm -f elf -F stabs hw5_1.asm ;;; ; ld -melf_i386 -o hw5_1 hw5_1.o ;;; ; ./hw5_1 ;;; ; ------------------------------------------------------------------- extern _printString extern _getInput extern _println ;;; ------------------------------------------------------------ ;;; data areas ;;; ------------------------------------------------------------ section .data prompt db "> " stars db "******************************************" N dd 0 saveECX dd 0 ;;; ------------------------------------------------------------ ;;; code area ;;; ------------------------------------------------------------ section .text global _start _start: ;;; get N from user mov ecx, prompt mov edx, 2 call _printString call _getInput mov dword[N], eax ;;; ------------------------------------------------- ;;; print opening triangle mov ecx, dword[N] for0: mov dword[saveECX], ecx mov edx, dword[N] sub edx, ecx inc edx mov ecx, stars call _printString call _println mov ecx, dword[saveECX] loop for0 ;;; ------------------------------------------------- ;;; print square mov ecx, dword[N] for1: mov dword[saveECX], ecx ;;; print a line of N stars mov ecx, stars mov edx, dword[N] call _printString call _println mov ecx, dword[saveECX] loop for1 ;;; ------------------------------------------------- ;;; print closing triangle mov ecx, dword[N] for2: mov dword[saveECX], ecx mov edx, ecx mov ecx, stars call _printString call _println mov ecx, dword[saveECX] loop for2 ;;; exit() mov eax,1 mov ebx,0 int 0x80 ; final system call
hw5_2.sh
#! /bin/bash # hw5_2.sh # D. Thiebaut # gets a number of the command line and prints a triangle # a square and a triangle with the width equal to the value # of the number. N=$1 for i in $( seq 1 $N ) ; do for j in $( seq 1 $i ) ; do echo -n "*" done echo "" done for i in $( seq 1 $N ) ;do for j in $( seq 1 $N ) ; do echo -n "*" done echo"" done exit for i in $( seq 1 $N ) ;do top=$( expr $N - $i + 1 ) for j in $( seq 1 $top ) ; do echo -n "*" done echo"" done
