CSC231 Homework 8 2015

--D. Thiebaut (talk) 12:40, 2 December 2015 (EST)

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Page under construction!

This is the last homework, and it is due on 12/14/15, the last day of class, at 11:55 p.m.

Problem 1

Write an assembly program called hw8.asm that computes the product of 2 integer numbers recursively, using only additions, as the algorithm below illustrates.

Algorithm

Inputs

a and b are integers

a is a positive integer and can be null

Algorithm

mult( a, b ) is recursive, and returns a multiplied by b

if a==0 mult( a, b ) = 0

if a==1 mult( a, b ) = b

otherwise mult( a, b ) = mult( a-1, b ) + b

Python Version

Your program should emulate the Python program below, at least in the way the mult() function operates.

<souce lang="python">

1. ! /usr/bin/python
2. mult.py
3. implements multiplication as a recursive
4. process. Works only with integers. a must
5. be positive.
6. mult( a, b ) = 0 if a == 0
7. mult( a, b ) = b if a == 1
8. = b + mult( a-1, b ) otherwise

def mult( a, b ):

   if a==0: return 0
   if a==1: return b
   return b + mult( a-1, b )

def main():

   a = 0
   b = 100
   print "a = %d  b = %d  mult( %d, %d ) = %d " \
         % ( a, b, a, b, mult( a, b ) )

   a = 1
   b = 100
   print "a = %d  b = %d  mult( %d, %d ) = %d " \
         % ( a, b, a, b, mult( a, b ) )

   a = 3
   b = 100
   print "a = %d  b = %d  mult( %d, %d ) = %d " \
         % ( a, b, a, b, mult( a, b ) )

   a = 100
   b = 0
   print "a = %d  b = %d  mult( %d, %d ) = %d " \
         % ( a, b, a, b, mult( a, b ) )

main()

</source>

Main Program

• Your main program should prompt the user for 2 integers, as illustrated below, output a blank line, and then output the result.
• The blank line is important, as it will be used to separate input from output.
• Example of user interaction:

./hw8
> 3
> 10

30

./hw8
> 5
> 11

55

<showafterdate after="20151215 00:00" before="20151230 00:00">

Solution Program

;;; ---------------------------------------------------------------------------
;;; mult.asm 
;;; D. Thiebaut
;;; computes the product of two positive integers a, and b, using
;;; a recursive method.
;;; ---------------------------------------------------------------------------
%include "asm_io.inc"

	section 	.data
a	dd		3
b	dd		100
	
	section		.text
	global		asm_main
;;; ---------------------------------------------------------------------------
asm_main:
	mov		eax, [ a ] 		; print a
	call		print_int
	call		print_nl
	
	mov		eax, [ b ] 		; print b
	call		print_int
	call		print_nl
	
	push		eax			; make room for returned value
	push		dword[ a ]              ; pass a
	push		dword[ b ]              ; pass b
	call		mult                    ; call mult( a, b )
	pop		eax                     ; get product
	call		print_int               ; print it
	call		print_nl
	call		print_nl
	ret
	
;;; ---------------------------------------------------------------------------
;;; mult( a, b )
;;; implements following algorithm:
;;; def mult( a, b ):
;;;     if a==0: return 0
;;;     if a==1: return b
;;;     return b + mult( a-1, b )
;;; 
;;;  stack after "mov ebp, esp"
;;;         ??     <---ebp+16
;;;          a     <---ebp+12
;;;          b     <---ebp+8
;;;       ret addr <---ebp+4
;;;       old ebp  <---ebp
;;; 
mult:	push		ebp
	mov		ebp, esp

%define m_result  dword[ebp+16]
%define m_a       dword[ebp+12]
%define m_b       dword[ebp+8]

.if1:	
	cmp		m_a, 0			; is a == 0 ?
	jne		.if2
	mov		m_result, 0 		; yes: return 0
	pop		ebp
	ret		2*4

.if2:	cmp		m_a, 1			; is a == 1 ?
	jne		.else2	
	mov		eax, m_b 		; yes: return b 
	mov		m_result, eax
	pop		ebp
	ret		2*4

.else2:	push		eax			; make room for value returned
	mov		eax, m_a		; by mult( )
	dec		eax
	push		eax			; pass a-1
	mov		eax, m_b
	push		eax			; pass b
	call		mult			; compute mult( a-1, b )
	pop		ebx			; ebx <-- (a-1)*b
	mov		eax, m_b		; eax <-- b
	add		eax, ebx		; eax <-- b+(a-1)*b
	mov		m_result, eax		; pass back
	
	pop		ebp
	ret		2*4
	

	
</source>
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