Difference between revisions of "CSC231 Homework Solutions 2017"

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--[[User:Thiebaut|D. Thiebaut]] ([[User talk:Thiebaut|talk]]) 17:57, 22 April 2017 (EDT)
 
--[[User:Thiebaut|D. Thiebaut]] ([[User talk:Thiebaut|talk]]) 17:57, 22 April 2017 (EDT)
 
----
 
----
 +
<onlydft>
 
=Problem 1=
 
=Problem 1=
 
::<source lang="asm">
 
::<source lang="asm">
Line 851: Line 852:
 
</source>
 
</source>
 
<!-- ========================================================= -->
 
<!-- ========================================================= -->
−
<onlydft>
+
<br />
 
=hw8_1.c=
 
=hw8_1.c=
 
<br />
 
<br />
Line 906: Line 907:
  
 
</source>
 
</source>
 +
=hw8_2.asm=
 +
<br />
 +
::<source lang="asm">
 +
; hw8_2.asm
 +
; D. Thiebaut
 +
 +
global f1
 +
global f2
 +
global f3
 +
 +
;;; ------------------------------------------------------
 +
;;; f1( s ): modifies s in place and make it uppercase.
 +
f1: push ebp
 +
mov ebp, esp
 +
push ebx
 +
mov ebx, dword[ebp+8]
 +
.for: cmp byte[ebx], 0
 +
je .done
 +
cmp byte[ebx],'a'
 +
jl .endfor
 +
cmp byte[ebx],'z'
 +
jg .endfor
 +
add byte[ebx],'A'-'a'
 +
.endfor:
 +
inc ebx
 +
jmp .for
 +
.done:
 +
pop ebx
 +
pop ebp
 +
ret 4
 +
 +
;;; ------------------------------------------------------
 +
;;; f2( a, b, c )
 +
;;; returns 2a + 3b -c in eax
 +
;;; computes it as 2(a+b) + b - c (fewer operations)
 +
f2: push ebp
 +
mov ebp, esp
 +
mov eax, dword[ebp+8+4+4] ; get first param
 +
add eax, dword[ebp+8+4]  ; eax <- a+b
 +
add eax, eax       ; eax <- 2(a+b)
 +
add eax, dword[ebp+8+4]  ; eax <- 2(a+b) + b
 +
sub eax, dword[ebp+8]    ; eax <- 2a + 3b -c
 +
pop ebp       ;
 +
ret 4*3       ; return and get rid of 3 ints
 +
 +
 +
;;; ------------------------------------------------------
 +
;;; f3( array, n )
 +
;;; computes the number of even ints in array, of length n.
 +
;;; returns this number in eax.
 +
f3: push ebp
 +
mov ebp, esp
 +
 +
;;; push ebx
 +
push ecx
 +
 +
mov ebx, dword[ebp+8+4] ; ebx <- array
 +
mov ecx, dword[ebp+8] ; ecx <- n
 +
mov eax, 0 ; number of even ints
 +
.for: and dword[ebx], 1 ; and array[i] with 1
 +
jnz .notEven ; if result is 1, then odd number
 +
.even: inc eax ; otherwise, add 1 to counter.
 +
.notEven:
 +
add ebx, 4
 +
loop .for
 +
 +
;;; pop ecx
 +
pop ebx
 +
 +
pop ebp
 +
ret 2*4
 +
 +
 +
</source>
 +
<!-- ================================================================= -->
 +
 +
=Homework 9: funcs.c=
 +
<br />
 +
::<source lang="c">
 +
#include <stdio.h>
 +
#include <stdlib.h>
 +
#include <string.h>
 +
 +
// getMax: returns the largest of 3 integers passed
 +
// by value
 +
int getMin( int a, int b, int c ) {
 +
  if ( a <= b && a <= c )
 +
    return a;
 +
  if ( b <= a && b <= c )
 +
    return b;
 +
  return c;
 +
}
 +
 +
// zap: given a string haystack, and a string needle,
 +
// looks for the firlst location of needle in the haystack, and
 +
// when it finds it, replaces it with dashes.  Example
 +
// char s1[] = "Mississippi";
 +
// char s2[] = "ss";
 +
// char *p  = s1, *q = s2;
 +
// zap( s1, s2 ) will replace s1 with "Mi--issippi".
 +
// zap( s1, "mm" ) will leave s1 unchanged.
 +
// Returns 0 if zap couldn't fine the needle in
 +
// the haystack.  Returns 1 if some character replacement
 +
// took place.
 +
int zap( char *haystack, char *needle ) {
 +
  char *p = strstr( haystack, needle );
 +
  char *q;
 +
  if ( p==NULL )
 +
    return 0;
 +
  for ( q=p; q != p+strlen( needle ); q++ )
 +
    *q = '-';
 +
  return 1;
 +
}
 +
 +
// merge(): takes two sorted arrays of ints of length 5
 +
// and merges them into an array of length 10, so that
 +
// the third array is sorted, in increasing order.
 +
void merge( int A[], int B[], int C[] ) {
 +
  int i = 0;
 +
  int j = 0;
 +
  int k = 0;
 +
  while ( 1 ) {
 +
    if ( A[i] < B[j] ) {
 +
      C[k++] = A[i++];
 +
      //printf( "A[%d] (%d) < B[%d] (%d) ==> C[%d] (%d)\n", i-1, A[i-1], j, B[j], k-1, C[k-1] );
 +
    }
 +
    else {
 +
      C[k++] = B[j++];
 +
      //printf( "A[%d] (%d) > B[%d] (%d) ==> C[%d] (%d)\n", i, A[i], j-1, B[j-1], k-1, C[k-1] );
 +
    }
 +
    if ( i>=5 || j>=5 )
 +
      break;
 +
  } 
  
 +
  while ( i<5 && k <10 ) {
 +
    C[k++] = A[i++];
 +
    //printf( "C[%d] (%d) <--  A[%d] (%d)\n", k-1, C[k-1], i-1, A[i-1] );
 +
  }
 +
  while ( j<5 && k <10 ) {
 +
    C[k++] = B[j++]; 
 +
    //printf( "C[%d] (%d) <--  B[%d] (%d)\n", k-1, C[k-1], j-1, B[j-1] );
 +
 +
  }
 +
}
 +
</source>
 +
=Homework 9: hanoi.asm=
 +
<br />
 +
::<source lang="asm">
 +
;;; hanoi.asm
 +
;;; D. Thiebaut 
 +
;;;
 +
;;; This program solves the "Towers of Hanoi" program
 +
;;; in assembly.  It prompts the user for an integer
 +
;;; number of disks (must be larger than 0) and displays
 +
;;; the name of the peg from which to move a disk, and
 +
;;; the name of the peg to move it to.  The pegs are
 +
;;; labeled 'A', 'B', and 'C', and the disks are always
 +
;;; assumed to moved originally from 'A' to 'B'.
 +
;;; To assemble, link, and run:
 +
;;;    nasm -f elf  231Lib.o
 +
;;;    nasm -f elf  hanoi.asm 
 +
;;;    ld -melf_i386 -o hanoi hanoi.o 231Lib.o
 +
;;;    ./hanoi  3   
 +
;;;     
 +
 +
                section .data
 +
N dd 5
 +
 +
                section .text
 +
extern _getInput
 +
extern _println
 +
extern _printString
 +
extern _atoi
 +
        global  _start
 +
_start:
 +
 +
;;; get N from command line
 +
;;; When any assembly language program starts, the operating system
 +
;;; passes it argc and argv through the statck.  The esp register
 +
;;; points to argc.  at esp+4, is a pointer to the beginning of argv[0],
 +
;;; as a string.  At esp+8 is a pointer to the beginning of argv[1],
 +
;;; as a string.
 +
 +
mov ebp, esp
 +
mov eax, dword[ebp]     ; put argc into eax
 +
;;; call _printDec     ; print it
 +
;;; call _println
 +
 +
mov eax, dword[ebp+4+4] ; make eax points to arv[1]
 +
call _atoi     ; convert ascii string to int
 +
mov dword[N], eax     ; save N
 +
 +
 +
;;; define the 3 pegs and pass them in bl, cl, and dl.
 +
 +
mov bl, 'A'
 +
mov cl, 'B'
 +
mov dl, 'C'
 +
 +
;;; moveDisks( N, 'A', 'B', 'C' )      ; eax <- N
 +
call moveDisks ; bl  <- 'A'
 +
; cl  <- 'B'
 +
; dl  <- 'C'
 +
 +
 +
;;; exit
 +
                mov    ebx, 0
 +
                mov    eax, 1
 +
                int    0x80
 +
 +
 +
;;; ------------------------------------------------------------------
 +
;;; moveDisks( n, source, dest, extra )
 +
;;;          eax  bl    cl    dl
 +
;;; Moves the n disks from source to dest using extra if necessary.
 +
;;; Uses recursion to move the N-1 disks above the last one.
 +
;;; Does not modify any of the registers
 +
;;; ------------------------------------------------------------------
 +
moveDisks: pushad
 +
 +
;;; if n==1:
 +
;;;    print( source, dest )
 +
cmp eax, 1
 +
jg recurse
 +
mov al, bl ; print source
 +
call printChar
 +
mov al, ' ' ; print space
 +
call printChar
 +
mov al, cl ; print dest
 +
call printChar
 +
call _println ; print \n
 +
 +
popad ; done! return
 +
ret
 +
 +
recurse:
 +
;;; moveDisks( n-1, source, temp, dest )
 +
dec eax ; eax <- n-1
 +
xchg cl, dl ; swap cl & dl
 +
call moveDisks      ; move n-1
 +
xchg cl, dl ; swap them back
 +
 +
 +
;;; print( source, dest )
 +
                mov    al, bl ; print source
 +
call    printChar
 +
mov    al, ' ' ; print space
 +
        call    printChar
 +
        mov    al, cl ; print dest
 +
        call    printChar
 +
        call    _println ; print \n
 +
 +
;;; moveDisks( n-1, temp, dest, source )
 +
                popad                      ; get all the original parameters back
 +
                pushad                      ; and push them back in the stack
 +
                xchg    bl, dl              ; exchange source and temp
 +
                dec    eax                ; eax <-- n-1
 +
                call    moveDisks          ; recurse
 +
 +
                popad                      ; pop all registers back
 +
                ret
 +
 +
;;; ------------------------------------------------------------------
 +
;;; printChar: prints the character in al
 +
;;; does not modify any other register
 +
;;; ------------------------------------------------------------------
 +
section .data
 +
char db 'A'
 +
section .text
 +
printChar: pushad
 +
mov byte[char],al
 +
mov ecx, char
 +
mov edx, 1
 +
call _printString
 +
popad
 +
ret
 +
 +
 +
 +
</source>
 
</onlydft>
 
</onlydft>
 
<br />
 
<br />

Latest revision as of 16:55, 16 September 2017

--D. Thiebaut (talk) 17:57, 22 April 2017 (EDT)


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