CSC231 Homework 7 Solution Programs

--D. Thiebaut 13:21, 1 November 2012 (EDT)

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Program 1

• This solution is from Gavi and shows she got the hints I gave the class about this array of character. The trick here is that since all the characters are listed in order in the string, you have character '0' at index 0 in the string. Character '1' at Index 1. Character '2' at Index 2... Character '9' at Index 9. And now for the magic: Character 'A' is at Index 10. Character 'B' at Index 11, all the way to Character 'F' at Index 15. So we do not need to test whether the number we want to print in hex is less than 10 or not. We just take the number, use it as an index in the string and pull the character that corresponds to it!

;;; ; hw7a.asm
;;; ; Gavi Levy Haskell
;;; ; 231a-ae
;;; ;
;;; ;
;;; ; Displays the contents of a double-word in decimal and
;;; ; hexadecimal.
;;; ;
;;; ; Prints:
;;; ;   x = 305441741 = 0x1234ABCD
;;; ;
;;; ;
;;; ;     nasm -f elf -F  stabs hw7a.asm
;;; ;     gcc -m32 -o hw7a driver.c asm_io.o hw7a.o
;;; ;     ./hw7a

%include "asm_io.inc"
	
	;; -----------------
	;; DATA SECTION
	;; -----------------
	section	.data
x	dd	0x1234ABCD
msg	dd	"  x = ", 0x00
msg2	dd	" = 0x", 0x00
hexChr	db	"0123456789ABCDEF"
	
	;; -----------------
	;; CODE SECTION
	;; -----------------
	section	.text
	global	asm_main
asm_main:
	mov	eax, msg
	call	print_string 	; print "  x = "
	
 	mov	eax, dword[x]
 	call	print_int	; print "305441741"

	mov	eax, msg2
	call	print_string	; print " = 0x"
	
 	mov	ecx, 8
	
for:	mov	ebx, dword[x]	; ebx <-- current rotation of x
 	rol	ebx, 4		; rotate first byte to end
 	mov	dword[x], ebx	; save current rotation in x
 	and	ebx, 0x000000F 	; keep only last digit in ebx

	mov	al, byte[hexChr + ebx]
	call	print_char	; print the character of hexChr
				; corresponding to the digit
	loop	for

	call	print_nl
	
	;; return to C program
	ret

Another good solution that doesn't use the array is provided by Naomi:

;;; hw7a.asm
;;; 231a-ad
;;; Naomi Long
;;;
;;; This program prints the decimal representation of the hex number
;;; 0x1234ABCD and then prints out the hex digits. The output is shown
;;; below, sandwiched between two prompts:
;;; 
;;; [231a-ad@grendel ~/hw7]$ ./hw7a
;;; x = 305441741 = 0x1234ABCD
;;; [231a-ad@grendel ~/hw7]$

	%include "asm_io.inc"

;;;  -------------------------
;;;  data segment
;;;  -------------------------

	section .data
	x	dd	0x1234ABCD
	xeq	db	"x = ",0x00
	eq	db	" = 0x",0x00
	
;;;  -------------------------
;;;  code area
;;;  -------------------------

	section .text
	global  asm_main
asm_main:
;;; print "x = "
	mov	eax, xeq
	call	print_string		; prints "x = "
	
;;; print decimal representation and " = 0x"
	mov	eax, dword[x]
	call	print_int		; prints 305441741
	mov	eax, eq			
	call	print_string		; prints " = 0x"
	
;;; print hex representation
	mov	ecx, 8			; 8 hex digits to loop over
start:
	mov	eax, dword[x]		; gets current hex rotation
	rol	eax, 4			; moves first hex to last position
	mov	dword[x], eax		; saves rotation
	and	eax, 0x0000000f		; remove all but last hex digit
	
	cmp	eax, 10			; check if hex digit letter or number
	jae	letr			; jump to treat letter as ascii char
	
	call	print_int		; prints last hex digit if number
	jmp 	loop			; skip to end of loop
	
letr:	add	eax, 'A'-10		; offset dec to ascii character
	call	print_char		; prints last hex number as ascii char

loop:	loop 	start			; loop to print all 8 hex digits

	call	print_nl		; prints new line after loop is done
	
;;;  return to C program
	ret