CSC231 Homework 8 Solution 2012
--D. Thiebaut 12:15, 13 November 2012 (EST)
Contents
Problem #1, Version 1
This version contains the complete code. Version 2 below separates the dumpRegs functions and main program into two files.
;;; ---------------------------------------------------------
;;; hw8aSol.asm
;;; D. Thiebaut
;;; 11/13/12
;;; This program supports several functions that allow the user
;;; to dump the contents of the main 6 registers (eax, ebx, ecx,
;;; edx, esi and edi) to the screen. Great for debugging without
;;; a debugger.
;;;
;;; Typical output:
;;; +-----------------+
;;; | eax = 1234 5678 |
;;; | ebx = 55FF 55FF |
;;; | ecx = FEDC BA98 |
;;; | edx = 0000 0000 |
;;; | esi = 1111 2222 |
;;; | edi = 2222 3333 |
;;; +-----------------+
;;; +-----------------+
;;; | eax = 1234 5678 |
;;; | ebx = 55FF 55FF |
;;; | ecx = FEDC BA98 |
;;; | edx = 0000 0000 |
;;; | esi = 1111 2222 |
;;; | edi = 2222 3333 |
;;; +-----------------+
;;;
;;; To assemble, link and run:
;;; nasm -f elf hw8aSol.asm
;;; ld -melf_i386 hw8aSol.o -o hw8aSol
;;; ./hw8aSol
;;; ---------------------------------------------------------
section .data
hex db "0123456789ABCDEF"
regs db "axbxcxdxsidi"
line db "+-----------------+", 10
string db "| exx = .... .... |", 10
lineLen equ $-string
reg equ string+3
word1 equ string+8
section .text
global _start
_start:
;;; Initialize the registers
mov eax, 0x12345678
mov ebx, 0x55FF55FF
mov ecx, 0xFEDCBA98
mov edx, 0x00000000
mov esi, 0x11112222
mov edi, 0x22223333
;;; dump them twice to verify that no registers gets modified...
call dumpRegs
call dumpRegs
;;; exit back to OS
mov ebx, 0
mov eax, 1
int 0x80
;;; ---------------------------------------------------------
;;; dumpRegs: dumps the contents of the 6 main registers (eax
;;; ebx, ecx, edx, esi and edi) in a box on the screen.
;;; the screen. Does not modify any register
;;; ---------------------------------------------------------
dumpRegs: pushad
;;; put all the registers to print in the stack. The loop
;;; will pop each one, one at a time, once per loop.
push edi
push esi
push edx
push ecx
push ebx
push eax
;;; print top bar
call printBar
;;; prepare to loop 6 times (6 registers )
mov ecx, 6
mov esi, 0
.for pop ebx ;get value of register to convert from stack
push ecx ;save loop counter
lea eax, [regs + esi*2] ;get address of string representing register
mov ax, word [eax] ;ax gets "ax" or "bx" or "cx" or ... "di"
mov word[reg], ax ;modify string with name of register
mov edx, word1 ;edx points to buffer where to store hex digits
;ebx contains register to convert
call storeDWord ;store hex equivalent of ebx in string
mov ecx, string ;print the whole string
mov edx, lineLen
call printInt80
inc esi ;index into register names incremented
pop ecx ;get loop counter back
loop .for
;;; print bottom bar
call printBar
popad
ret
;;; ---------------------------------------------------------
;;; storeDWord: gets
;;; ebx: register to convert
;;; edx: address where to put information
;;; ---------------------------------------------------------
storeDWord: rol ebx, 16 ;bring most sig word in least sig word
call storeWord
inc edx ;make edx skip space in string
rol ebx, 16 ;get lower byte back in least sig pos
call storeWord
ret
;;; ---------------------------------------------------------
;;; storeWord: gets
;;; bx: word to convert
;;; edx: address where to put information. edx incremented by 4
;;; ---------------------------------------------------------
storeWord: rol bx, 8 ;put most sig byte of bx in lower position
call storeByte ;and convert it. Store it in where edx
;points to.
rol bx, 8 ;same with lower nybble
call storeByte
ret
;;; ---------------------------------------------------------
;;; storeByte: gets
;;; bl: byte to convert
;;; edx: address where to store information. edx incremented by 2 automatically
;;; ---------------------------------------------------------
storeByte: push ebx
push eax
rol bl, 4 ;get upper nybble in lower position (LSB)
mov al, bl ;play with al now
and eax, 0x0000000F ;clear all but the least significant nybble
add eax, hex ;get ascii in hex array equivalentn to nybble
mov al, [eax] ;hex digit now in bl
mov byte [edx], al ;store ascii in string
inc edx
rol bl, 4 ;get lower nybble back
mov al, bl ;play with al
and eax, 0x0000000F
add eax, hex
mov al, [eax]
mov byte [edx], al
inc edx
pop eax
pop ebx
ret
;;; ---------------------------------------------------------
;;; printInt80: prints on the screen the string whose address
;;; is in ecx and length in edx.
;;; does not modify registers
;;; ---------------------------------------------------------
printInt80: push eax
push ebx
mov eax, 4
mov ebx, 1
int 0x80
pop ebx
pop eax
ret
;;; ---------------------------------------------------------
;;; printBar: prints the horizontal bar
;;; does not modify registers
;;; ---------------------------------------------------------
printBar: push ecx
push edx
mov ecx, line
mov edx, lineLen
call printInt80
pop edx
pop ecx
ret
Problem #1, Version 2
Main Program
;;; ---------------------------------------------------------
;;; hw8aSol2.asm
;;; D. Thiebaut
;;; 11/13/12
;;; This program uses the library dumpRegs.asm to dump
;;; the contents of the main 6 registers (eax, ebx, ecx,
;;; edx, esi and edi) to the screen. Great for debugging without
;;; a debugger.
;;;
;;; Typical output:
;;; +-----------------+
;;; | eax = 1234 5678 |
;;; | ebx = 55FF 55FF |
;;; | ecx = FEDC BA98 |
;;; | edx = 0000 0000 |
;;; | esi = 1111 2222 |
;;; | edi = 2222 3333 |
;;; +-----------------+
;;; +-----------------+
;;; | eax = 1234 5678 |
;;; | ebx = 55FF 55FF |
;;; | ecx = FEDC BA98 |
;;; | edx = 0000 0000 |
;;; | esi = 1111 2222 |
;;; | edi = 2222 3333 |
;;; +-----------------+
;;;
;;; To assemble, link and run:
;;; nasm -f elf hw8aSol.asm
;;; ld -melf_i386 hw8aSol.o -o hw8aSol
;;; ./hw8aSol
;;; ---------------------------------------------------------
%include "dumpRegs.asm"
section .text
global _start
_start:
;;; Initialize the registers
mov eax, 0x12345678
mov ebx, 0x55FF55FF
mov ecx, 0xFEDCBA98
mov edx, 0x00000000
mov esi, 0x11112222
mov edi, 0x22223333
;;; dump them twice to verify that no registers gets modified...
call dumpRegs
call dumpRegs
;;; exit back to OS
mov ebx, 0
mov eax, 1
int 0x80
Library
;;; ---------------------------------------------------------
;;; dumpRegs.asm
;;; D. Thiebaut
;;; 11/13/12
;;; This libary supports several functions that allow the user
;;; to dump the contents of the main 6 registers (eax, ebx, ecx,
;;; edx, esi and edi) to the screen. Great for debugging without
;;; a debugger.
;;;
;;; Typical output:
;;; +-----------------+
;;; | eax = 1234 5678 |
;;; | ebx = 55FF 55FF |
;;; | ecx = FEDC BA98 |
;;; | edx = 0000 0000 |
;;; | esi = 1111 2222 |
;;; | edi = 2222 3333 |
;;; +-----------------+
;;; Include in program as follows:
;;;
;;;
;;; %include "dumpRegs.asm"
;;; ---------------------------------------------------------
section .data
hex db "0123456789ABCDEF"
regs db "axbxcxdxsidi"
line db "+-----------------+", 10
string db "| exx = .... .... |", 10
lineLen equ $-string
reg equ string+3
word1 equ string+8
section .text
;;; ---------------------------------------------------------
;;; dumpRegs: dumps the contents of the 6 main registers (eax
;;; ebx, ecx, edx, esi and edi) in a box on the screen.
;;; the screen. Does not modify any register
;;; ---------------------------------------------------------
dumpRegs: pushad
;;; put all the registers to print in the stack. The loop
;;; will pop each one, one at a time, once per loop.
push edi
push esi
push edx
push ecx
push ebx
push eax
;;; print top bar
call printBar
;;; prepare to loop 6 times (6 registers )
mov ecx, 6
mov esi, 0
.for pop ebx ;get value of register to convert from stack
push ecx ;save loop counter
lea eax, [regs + esi*2] ;get address of string representing register
mov ax, word [eax] ;ax gets "ax" or "bx" or "cx" or ... "di"
mov word[reg], ax ;modify string with name of register
mov edx, word1 ;edx points to buffer where to store hex digits
;ebx contains register to convert
call storeDWord ;store hex equivalent of ebx in string
mov ecx, string ;print the whole string
mov edx, lineLen
call printInt80
inc esi ;index into register names incremented
pop ecx ;get loop counter back
loop .for
;;; print bottom bar
call printBar
popad
ret
;;; ---------------------------------------------------------
;;; storeDWord: gets
;;; ebx: register to convert
;;; edx: address where to put information
;;; ---------------------------------------------------------
storeDWord: rol ebx, 16 ;bring most sig word in least sig word
call storeWord
inc edx ;make edx skip space in string
rol ebx, 16 ;get lower byte back in least sig pos
call storeWord
ret
;;; ---------------------------------------------------------
;;; storeWord: gets
;;; bx: word to convert
;;; edx: address where to put information. edx incremented by 4
;;; ---------------------------------------------------------
storeWord: rol bx, 8 ;put most sig byte of bx in lower position
call storeByte ;and convert it. Store it in where edx
;points to.
rol bx, 8 ;same with lower nybble
call storeByte
ret
;;; ---------------------------------------------------------
;;; storeByte: gets
;;; bl: byte to convert
;;; edx: address where to store information. edx incremented by 2 automatically
;;; ---------------------------------------------------------
storeByte: push ebx
push eax
rol bl, 4 ;get upper nybble in lower position (LSB)
mov al, bl ;play with al now
and eax, 0x0000000F ;clear all but the least significant nybble
add eax, hex ;get ascii in hex array equivalentn to nybble
mov al, [eax] ;hex digit now in bl
mov byte [edx], al ;store ascii in string
inc edx
rol bl, 4 ;get lower nybble back
mov al, bl ;play with al
and eax, 0x0000000F
add eax, hex
mov al, [eax]
mov byte [edx], al
inc edx
pop eax
pop ebx
ret
;;; ---------------------------------------------------------
;;; printInt80: prints on the screen the string whose address
;;; is in ecx and length in edx.
;;; does not modify registers
;;; ---------------------------------------------------------
printInt80: push eax
push ebx
mov eax, 4
mov ebx, 1
int 0x80
pop ebx
pop eax
ret
;;; ---------------------------------------------------------
;;; printBar: prints the horizontal bar
;;; does not modify registers
;;; ---------------------------------------------------------
printBar: push ecx
push edx
mov ecx, line
mov edx, lineLen
call printInt80
pop edx
pop ecx
ret
Problem #2
;;; ; hw8b.asm
;;; ; Gavi Levy Haskell
;;; ; 231a-ae
;;; ;
;;; ; Uses functions to print the first ten
;;; ; lines of the Pascal Triangle in hex
;;; ;
;;; ; prints:
;;; ;
;;; ; 01 00 00 00 00 00 00 00 00 00
;;; ; 01 01 00 00 00 00 00 00 00 00
;;; ; 01 02 01 00 00 00 00 00 00 00
;;; ; 01 03 03 01 00 00 00 00 00 00
;;; ; 01 04 06 04 01 00 00 00 00 00
;;; ; 01 05 0A 0A 05 01 00 00 00 00
;;; ; 01 06 0F 14 0F 06 01 00 00 00
;;; ; 01 07 15 23 23 15 07 01 00 00
;;; ; 01 08 1C 38 46 38 1C 08 01 00
;;; ; 01 09 24 54 7E 7E 54 24 09 01
;;; ;
;;; ;
;;; ; nasm -f elf -F stabs hw8b.asm
;;; ; ld -melf_i386 -o hw8b hw8b.o
;;; ; ./hw8b
;; -----------------
;; DATA SECTION
;; -----------------
section .data
Pascal times 10 db 0
hexChr db "0123456789ABCDEF"
return db 10
space db " "
;; -----------------
;; CODE SECTION
;; -----------------
section .text
global _start
_start:
mov ebx, Pascal ; pass address of array in ebx
call init ; store 0 in Pascal array and 1
; in first cell
mov ecx, 10
for: mov ebx, Pascal ; pass address of array
call printArray ; print Pascal array
mov ebx, Pascal
call nextLine ; compute next line of triangle
loop for
;;; exit
mov eax, 1
mov ebx, 0
int 0x80
;;; init
;;; recieves address of Pascal in ebx
;;; modifies no registers
init:
mov dword[ebx], 0 ; change first 4 terms to zero
mov dword[ebx + 4], 0 ; change next 4 terms to 0
mov word[ebx + 8], 0 ; change last 2 terms to 0
mov byte[ebx], 1 ; change first term of array to 1
ret
;;; printArray
;;; recieves address of Pascal in ebx
;;; modifies eax, ebx, edx
printArray:
push ecx ; save outer loop position
mov ecx, 10
.for:
push ecx ; save loop position
push ebx ; save address
mov cl, byte[ebx] ; first digit
ror ecx, 4 ; get correct digit
and ecx, 0x0F ; keep only this digit
add ecx, hexChr ; add address of hexChr
mov eax, 4
mov ebx, 1
mov edx, 1
int 0x80 ; print first digit
pop ebx
push ebx
mov cl, byte[ebx] ; second digit
and ecx, 0x0F ; keep only this digit
add ecx, hexChr ; add address of hexChr
mov eax, 4
mov ebx, 1
mov edx, 1
int 0x80 ; print second digit
mov eax, 4
mov ebx, 1
mov ecx, space
mov edx, 1
int 0x80 ; print space
pop ebx
pop ecx ; retrieve loop position
add ebx, 1
loop .for ; retrieve outer loop position
mov eax, 4
mov ebx, 1
mov ecx, return ; return
mov edx, 1
int 0x80 ; next line
pop ecx
ret
;;; nextLine
;;; recieves address of Pascal in ebx
;;; modifies eax, ebx
nextLine:
push ecx ; save outer loop position
mov ecx, 9
add ebx, 9
.for:
push ecx ; save loop position
mov al, byte[ebx - 1] ; n = row[x - 1]
add byte[ebx], al ; row[x] += n
dec ebx ; x -= 1
pop ecx ; retrieve loop position
loop .for
pop ecx ; retrieve loop position
ret
Problem 3: Optional and Extra Credit
Option 1
- First option: write a program that constantly pushes stuff in the stack until it seg-faults. Make it print the size of the stack as it do so.
;;; hwtest.asm
;;; a simple demo program to test I/O with
;;; a C "wrapper" program
%include "asm_io.inc"
;; -------------------------
;; data segment
;; -------------------------
section .data
msg db "Hello! Please enter an integer number: ",0x00
msg2 db "You have entered ",0x00
x dd 0
;; -------------------------
;; code area
;; -------------------------
section .text
global asm_main
asm_main:
mov ebx, esp
for: mov eax, esp
sub eax, ebp
neg eax
call print_int
call print_nl
sub esp, 100000 ;equivalent to pushing 100,000 bytes
jmp for
ret
