CSC231 Homework 4 Solution

From dftwiki3
Revision as of 11:33, 14 October 2008 by Thiebaut (talk | contribs) (New page: =Homework #4 Solution= ==Problem #1== <code><pre> ;;; ; ------------------------------------------------------------------- ;;; ; hw4a.asm ;;; ; Lei Lei ;;; ; ;;; ; Given the following de...)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

Homework #4 Solution

Problem #1

;;; ; -------------------------------------------------------------------
;;; ; hw4a.asm
;;; ; Lei Lei
;;; ;
;;; ; Given the following definition:
;;; ; msg     	db    	3,"ALL",11,"PROGRAMMERS",3,"ARE",11,"PLAYWRIGHTS"
;;; ;         	db    	3,"AND",3,"ALL",9,"COMPUTERS",3,"ARE",5,"LOUSY"
;;; ;         	db    	6,"ACTORS"
;;; ; This program outputs each letter of msg capitalized in a line:
;;; ;
;;; ;  All
;;; ;  Programmers
;;; ;  Are
;;; ;  Playwrights
;;; ;  And
;;; ;  All
;;; ;  Computers
;;; ;  Are
;;; ;  Lousy
;;; ;  Actors
;;; ;
;;; ; to assemble and run:
;;; ;
;;; ;     nasm -f elf -F  stabs hw4a.asm
;;; ;     ld -o hw4a hw4a.o
;;; ;     ./hw4a
;;; ; -------------------------------------------------------------------


;;; ; -----------------------------------------------------------
;;; ; Constants
;;; ; -----------------------------------------------------------
	EXIT    equ             1
	WRITE   equ             4
	STDOUT  equ             1

;;;  ------------------------------------------------------------
;;;  data areas
;;;  ------------------------------------------------------------

	section	.data

msg     	db    	3,"ALL",11,"PROGRAMMERS",3,"ARE",11,"PLAYWRIGHTS"
        	db    	3,"AND",3,"ALL",9,"COMPUTERS",3,"ARE",5,"LOUSY"
        	db    	6,"ACTORS"
	
N       	equ   	10

newLine		db	0x0A
line_len	equ	1
	
counter		dd	0
temp1		dd	0
temp2		dd	0
	
;;;  ------------------------------------------------------------
;;;  code area
;;;  ------------------------------------------------------------
	section .text
	global	_start
	
_start:	mov	ecx, N
	
for:	mov	dword[temp1], ecx 	;Main loop, loop for N=10 times
	mov	esi, msg
 	add 	esi, dword[counter] 	;esi points to integer
	inc	dword[counter]

	mov	eax, WRITE
	mov	ebx, STDOUT
	mov	ecx, msg	
	add	ecx, dword[counter] 	;ecx points to first letter
	mov	edx, 1
 	int	0x80			;print first letter in cap
	
	movzx	ecx, byte[byte esi] 	;move the integer to ecx
	dec	ecx			;minus the already printed first letter
	inc	dword[counter]

    	call	printLower
 	call	printLine

 	mov	ecx, dword[temp1]
 	loop	for
	
;;;  exit()

	mov	eax,EXIT
	mov	ebx,0
	int	0x80			; final system call

;;; ----------------------------------------------------------------
;;; printLower: Print letters in lower case
;;; ----------------------------------------------------------------
printLower:
forLower:	
	mov	dword[temp2], ecx
	mov	eax, WRITE
	mov	ebx, STDOUT

	mov	ecx, msg
	add	ecx, dword[counter]
	add	dword[ecx], 'a'-'A'		;change into lower case
	mov	edx, 1			
	int 	0x80		

	inc	dword[counter]		
	mov	ecx, dword[temp2]	
	loop	forLower
	ret
	
;;; ---------------------------------------------------------------
;;; printLine: Hard break and print new line
;;; ---------------------------------------------------------------
printLine:
	section	.bss
.temp	resd	1

	section .text
	mov	eax, WRITE
	mov	ebx, STDOUT
	mov	edx, line_len
	mov	dword[.temp], ecx
	lea	ecx, [newLine]
	int	0x80
	mov	ecx, dword[.temp]
	ret
	


Problem #2

;;; ; hw4b.asm
;;; ; Julia Burch
;;; ; 231a-aa
;;; ;
;;; ; OUTPUT:
;;; ;  1	0	0	0	0	0	0	0	0	0
;;; ;  1	1	0	0	0	0	0	0	0	0
;;; ;  1	2	1	0	0	0	0	0	0	0
;;; ;  1	3	3	1	0	0	0	0	0	0
;;; ;  1	4	6	4	1	0	0	0	0	0
;;; ;  1	5	10	10	5	1	0	0	0	0
;;; ;  1	6	15	20	15	6	1	0	0	0
;;; ;  1	7	21	35	35	21	7	1	0	0
;;; ;  1	8	28	56	70	56	28	8	1	0
;;; ;  1	9	36	84	126	126	84	36	9	1
;;; ; to assemble and run:
;;; ;
;;; ;     nasm -f elf -F  stabs hw4b.asm
;;; ;     gcc -o hw4b driver.c asm_io.o hw4b.o
;;; ;     ./hw4b
;;; ; -------------------------------------------------------------------

	;; %include files here...
%include "asm_io.inc"

;;;  ------------------------------------------------------------
;;;  data areas
;;;  ------------------------------------------------------------

	section	.data

	shift	 db	0
	pascal 	 db 	1, 0, 0, 0, 0, 0, 0, 0, 0, 0
	temp	 dd     0
	space	 db	9
	rows	 equ	10
;;;  ------------------------------------------------------------
;;;  code area
;;;  ------------------------------------------------------------

	section	.text
	global	asm_main

asm_main:	

	;; move first line of Pascal into edi, then rows into ecx
	;; to initialize and run for loop 10 times.
	mov	edi, pascal
	mov	ecx, rows
main:
	;; store ecx in a temporary variable, move rows into ecx to initialize
	;; and run print_triangle loop. Move the first line into edi
	;; and 0 into esi.
	mov	dword[temp], ecx
	mov	ecx, rows
	mov	edi, pascal
	mov	esi, 0

print_triangle:
	;; print the int in al, print tab and then clear
	;; tab from register.
	mov	al, byte[edi+esi]
	call	print_int
	mov	eax, space
	call	print_string
	mov	eax, 0
	inc 	esi

	loop print_triangle

	;; print a new line
	call	print_nl

	;; load newly calculated pascal line into edi, then shift the row
	mov	edi, pascal
	mov	esi, shift
	mov	ecx, rows-1

calculate:
	;; move pascal digit into al, add it to previous row's, store the value
	mov	al, byte[edi+ecx]
	add	al, byte[esi+ecx]
	mov	byte[edi+ecx], al

	loop calculate

	;; restore value of ecx for loop
	mov ecx, dword[temp]
	loop main

;;;  exit()
	ret