CSC231 Homework 7 Solution 2010

--D. Thiebaut 15:04, 19 November 2010 (UTC)

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Part 1

hw7.txt

RB Axtell and 	Amy Tayloe
231a-af		231a-ai
10/11/2010

1) It would print ecx lines. The code has ecx as 0 so it will loop 256 times. 0 decrements to -1 which in signed numbers is 255.

2) Each time through the loop 4 bytes are added the stack. If the stack has 1000 bytes, after 251 (1000/4 + 1) lines of "hello world" are printed a segmentation fault will be thrown before the data section is overwritten. It will not run to completion (256 lines printed), but will stop 5 lines short of the end.

3) If standard error is sent to the same paper, the words "Segmentation fault" will be printed after 251 lines. Otherwise, there will be no evidence of the bug without counting the number of lines.

4) A quarter of the stack plus 1 or 256 if the stack is big enough (1024 bytes).

Part 2

;;; --------------------------------------------------------
;;; hw7.asm
;;; 
;;; RB Axtell
;;; 231a-af
;;; 10/11/2010
;;;
;;; Prints the last N digits of the integer x (uses leading
;;; 0's if shorter than N)
;;;
;;;
;;; I had a function sepAndPush that broke off one integer
;;; at a time and pushed it to the stack, but had problems
;;; because data was being pushed onto the stack after the
;;; return address. So it tried to use the last doubleword
;;; integer from x pushed to the stack as the ret address.
;;; 
;;; To compile and run:
;;;	nasm -f elf -F stabs hw7.asm
;;;	ld -melf_i386 -o hw7 hw7.o
;;;	./hw7
;;; --------------------------------------------------------

EXIT	equ	1
READ	equ	3
WRITE	equ	4
STDOUT	equ	1

	;; ------------------------------------------------
	;; data area
	;; ------------------------------------------------

	section .data

x	dd	1234589		; number to print
N	dd	10		; length of number to print
tmp	dd	0		; to store the current int

	;; ------------------------------------------------
	;; code area
	;; ------------------------------------------------

	section .text
	global	_start

_start:
	nop			;no operation for debugging
	nop
	nop
	mov	ecx, dword[N]	;number of digits to print
	mov	eax, dword[x]	;the number to print
	mov	ebx, 10

break:	xor	edx, edx
	div	ebx		;divide the number by 10
	push	edx		;the remainder contains the last digit
	loop	break		;end break

	mov	ecx, [N]	;reset the counter
	
print:	mov	esi, ecx	;store the counter
	pop	ecx		;pop the next digit
	call	printChar
	mov	ecx, esi	;re-store the counter
	loop	print		;end print

	;; Print a line feed
	mov	dword[tmp], 0x0a;line feed
	mov	eax, WRITE
	mov	ebx, STDOUT
	mov	ecx, tmp
	mov	edx, 1
	int	0x80
	
	jmp	theEnd

;;; ------------ printChar -------------
;;; digit to print in ecx
;;;
;;; converts the number in ecx to ASCII
;;; and prints.
;;; ------------------------------------
	
printChar:
	
	mov	dword[tmp], ecx	;tmp < digit to print
	add	dword[tmp], 0x30;+0x30 = change to ASCII
	mov	eax, WRITE
	mov	ebx, STDOUT
	mov	ecx, tmp	;address of digit in memory
	mov	edx, 1
	int	0x80
	
	ret			;end printChar

	;; exit()

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