Difference between revisions of "CSC231 Homework 8 2015"

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Your program should emulate the Python program below, at least in the way the '''mult()''' function operates.
 
Your program should emulate the Python program below, at least in the way the '''mult()''' function operates.
::<souce lang="python">
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::<source lang="python">
 
#! /usr/bin/python
 
#! /usr/bin/python
 
# mult.py
 
# mult.py
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==Main Program==
 
==Main Program==
 
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Revision as of 20:40, 2 December 2015

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


Page under construction!

UnderConstruction.jpg


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.

#! /usr/bin/python
# mult.py
# implements multiplication as a recursive
# process.  Works only with integers. a must
# be positive.
# mult( a, b ) = 0    if a == 0
# mult( a, b ) = b    if a == 1 
#              = 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()


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>
</showafterdate>