CSC231 Homework 1 Solutions 2010

--D. Thiebaut 13:52, 28 September 2010 (UTC)

---

hw1a.asm

;;; -------------------------------------------------------------------
;;; hw1a.asm
;;; 
;;; RB Axtell
;;; CSC 231
;;; 18/9/2010
;;;
;;; Size of executable: 1306 bytes measured with the command
;;;
;;;     echo `ls -l hw1a  | cut -d' ' -f 5` bytes
;;; 
;;; 
;;; Homework 1, Problem 1 :
;;;
;;; 	Prints :
;;; 
;;; abcdefghiklmnopqrstuvwxyz
;;; abcdefghiklmnopqrstuvw
;;; abcdefghiklmnopqrst
;;; abcdefghiklmnopq
;;; abcdefghiklmn
;;; abcdefghik
;;; abcdefg
;;; abcd
;;; a
;;; 
;;;
;;; to assemble and run:
;;;
;;;     nasm -f elf -F  stabs hw1a.asm
;;;     ld -melf_i386 -o hw1a hw1a.o
;;;     ./hw1a
;;; 
;;; -------------------------------------------------------------------
 
;%include files here...
 
EXIT    equ             1
WRITE   equ             4
STDOUT  equ             1
 
      	;; ------------------------------------------------------------
	;; data area
	;; ------------------------------------------------------------
 
	section	.data
 msg	db	"abcdefghijklmnopqrstuzwxyz",10
	db	"abcdefghijklmnopqrstuzw",10
	db	"abcdefghijklmnopqrst",10
	db	"abcdefghijklmnopq",10
	db	"abcdefghijklmn",10
	db	"abcdefghijk",10
	db	"abcdefgh",10
	db	"abcde",10
	db	"a",10
MSGLEN	equ	$-msg
	
	;; ------------------------------------------------------------
	;; code area
	;; ------------------------------------------------------------
 
	section	.text
	global	_start
 
_start: mov	eax, WRITE	; all nine lines
	mov	ebx, STDOUT
	mov	ecx, msg
	mov	edx, MSGLEN
	int	0x80
 
 	;; exit()
 
	mov	eax,EXIT
	mov	ebx,0
	int	0x80		; final system call

hw1b.asm

;;; -------------------------------------------------------------------
;;; hw1b.asm
;;; 
;;; RB Axtell
;;; CSC 231
;;; 18/9/2010
;;;
;;; Size of executable: 3632 bytes measured with the command
;;;
;;;     echo `ls -l hw1b  | cut -d' ' -f 5` bytes
;;; 
;;;	The B version of the program is almost three times as large as
;;; the A version. B uses 18 blocks to print; 1 for each line and one
;;; for each line break while A only uses 1. B doesn't have as much in
;;; the data section as A, but it takes so many more calls to print it
;;; all out,that it isn't worth the space saved in the data section.
;;;
;;; Homework 1, Problem 2 :
;;;
;;; 	Prints :
;;; 
;;; abcdefghiklmnopqrstuvwxyz
;;; abcdefghiklmnopqrstuvw
;;; abcdefghiklmnopqrst
;;; abcdefghiklmnopq
;;; abcdefghiklmn
;;; abcdefghik
;;; abcdefg
;;; abcd
;;; a
;;; 
;;;
;;; to assemble and run:
;;;
;;;     nasm -f elf -F  stabs hw1b.asm
;;;     ld -melf_i386 -o hw1b hw1b.o
;;;     ./hw1b
;;; 
;;; -------------------------------------------------------------------
 
;%include files here...
 
EXIT    equ             1
WRITE   equ             4
STDOUT  equ             1
 
      	;; ------------------------------------------------------------
	;; data area
	;; ------------------------------------------------------------
 
	section	.data
msg	db	"abcdefghijklmnopqrstuvwxyz"
MSGLEN	equ	$-msg
newln	db	10
	
	;; ------------------------------------------------------------
	;; code area
	;; ------------------------------------------------------------
 
	section	.text
	global	_start
 
_start:	mov	eax, WRITE	; print abcdefghiklmnopqrstuvwxyz
	mov	ebx, STDOUT
	mov	ecx, msg
	mov	edx, MSGLEN
	int	0x80
	
	mov	eax, WRITE	; line break
	mov	ebx, STDOUT
	mov	ecx, newln
	mov	edx, 1
	int	0x80

	mov	eax, WRITE	; abcdefghiklmnopqrstuvw
	mov	ebx, STDOUT
	mov	ecx, msg
	mov	edx, MSGLEN-3	; length of alphabet minus 3
	int	0x80
 
 	mov	eax, WRITE	; line break
	mov	ebx, STDOUT
	mov	ecx, newln
	mov	edx, 1
	int	0x80

	mov	eax, WRITE	; abcdefghiklmnopqrst
	mov	ebx, STDOUT
	mov	ecx, msg
	mov	edx, MSGLEN-6	; length of alphabet minus 6
	int	0x80

	mov	eax, WRITE	; line break
	mov	ebx, STDOUT
	mov	ecx, newln
	mov	edx, 1
	int	0x80

	mov	eax, WRITE	; abcdefghiklmnopq
	mov	ebx, STDOUT
	mov	ecx, msg
	mov	edx, MSGLEN-9
	int	0x80

	mov	eax, WRITE	; line break
	mov	ebx, STDOUT
	mov	ecx, newln
	mov	edx, 1
	int	0x80

	mov	eax, WRITE	; abcdefghiklmn
	mov	ebx, STDOUT
	mov	ecx, msg
	mov	edx, MSGLEN-12
	int	0x80

	mov	eax, WRITE	; line break
	mov	ebx, STDOUT
	mov	ecx, newln
	mov	edx, 1
	int	0x80

	mov	eax, WRITE	; abcdefghik
	mov	ebx, STDOUT
	mov	ecx, msg
	mov	edx, MSGLEN-15
	int	0x80

	mov	eax, WRITE	; line break
	mov	ebx, STDOUT
	mov	ecx, newln
	mov	edx, 1
	int	0x80

	mov	eax, WRITE	; abcdefg
	mov	ebx, STDOUT	
	mov	ecx, msg
	mov	edx, MSGLEN-18
	int	0x80

	mov	eax, WRITE	; line break
	mov	ebx, STDOUT
	mov	ecx, newln
	mov	edx, 1
	int	0x80

	mov	eax, WRITE	; abcd
	mov	ebx, STDOUT
	mov	ecx, msg
	mov	edx, MSGLEN-21
	int	0x80

	mov	eax, WRITE	; line break
	mov	ebx, STDOUT
	mov	ecx, newln
	mov	edx, 1
	int	0x80

	mov	eax, WRITE	; print "a", the final line
	mov	ebx, STDOUT
	mov	ecx, msg
	mov	edx, MSGLEN-25
 	int	0x80

 	mov	eax, WRITE	; line break
 	mov	ebx, STDOUT
	mov	ecx, newln
	mov	edx, 1
	int	0x80
	
	;; exit()
 
	mov	eax,EXIT
	mov	ebx,0
	int	0x80		; final system call