Assembly Language

i have changed my first PHPpage interface but it seem err not like 
what i want... it's look like so typical web. maybe im too new about 
this... have to learn and read more and more...

ok leave my PHP for a while ...i will explain in my next post bout 
that.. hehe

now. i want to do 3 task with assembly language.
this is a quite famaous questions to google for.. believe me. :P

no. im not doing this in my nano. im doing this coding using windows. 
(Microsoft c++ 2010 Express). im just try because i dont know wth of 
output i can get from the code. Still blur-ing.

6. Fibonacci Numbers

   Write a program that uses a loop to calculate the first seven values
   of the Fibonaccinumber sequence,described by the following formula:
   Fib(1) = 1, Fib(2) = 1, Fib(n) = Fib(n - 1) +   Fib(n - 2). Place
   each value in the EAX register and display it with a call 
   DumpRegsstatement (see Section 3.2) inside the loop. 


TITLE Question6 assignment1  (main.asm)

; Description:
; This progam adds and subtracts 32-bit unsigned
; integers and stores the sum in a variable
; Revision date:

INCLUDE Irvine32.inc

main PROC
 mov   eax,1
 call DumpRegs ;1st value    
 call DumpRegs ;2nd value

 mov   ebx,0 ; initial setup
 mov   edx,1
    mov   ecx,6 ; count

L1:
 mov  eax,ebx ; eax = ebx + edx
 add  eax,edx
 call DumpRegs ; display eax
 mov  ebx,edx
 mov  edx,eax
    Loop L1

 exit
main ENDP
END main


This is the output that i get ::




noted :: this is the wrong answer. why .? because the question want
         the first seven values... but it show you 8 in that output. 

The answer should be like this. pic below. But the coding must change
a bit, you cant start the number from = 0, but = 1 .




so the answer for this task is :: 110 1100


7. Arithmetic Expression

   Write a program that implements the following arithmetic 
   expression: EAX = −val2 + 7 − val3 + val1
   
   Use the following data definitions:
   val1 SDWORD 8
   val2 SDWORD  -15
   val3 SDWORD 20
   In comments next to each instruction, write the hexadecimal 
   value of EAX. Insert a  call
   
   DumpRegsstatement at the end of the program.


TITLE Question7 assignment1  (main.asm)

; Description:
; This progam adds and subtracts 32-bit unsigned
; integers and stores the sum in a variable
; Revision date:

INCLUDE Irvine32.inc

.data
 val1  SDWORd 8
 val2  SDWORD -15
 val3  SDWORD 20
 finalVal SDWORD ?

.code
main PROC

 mov eax, val2 
 neg eax ; eax=-15
 add eax, 7; -val2 + 7

 mov ebx, val3
 add ebx, val1 ; val3+val1

 sub eax,ebx

 mov finalVal, eax 
 call DumpRegs  ; display the registers

exit

main ENDP
END main


This is the output that i get ::




** double checking := −val2 + 7 − val3 + val1

                    = 15 + 7 - 20 + 8

                    = 10


8. Copy a String Backwards

   Write a program using the LOOP instruction with indirect addressing
   that copies a  string from source to target, reversing the 
   character order in the process. Use the following variables:
   source  BYTE "This is the source string",0
   target  BYTE  SIZEOF source DUP('#')

   Insert the following statements immediately after the loop to
   display the hexadecimal contents of the target string:
   
   movesi,OFFSET target ; offset of variable
   mov  ebx,1 ; byte format
   movecx,SIZEOF target ; counter
   callDumpMem

   If your program works correctly, it will display the following 
   sequence of hexadecimal bytes: 
   
   67 6E 69 72 74 73 20 65 63 72 75 6F 73 20 65 68 
   74 20 73 69 20 73 69 68 54


TITLE Question8 assignment1  (main.asm)

; Description:
; This progam adds and subtracts 32-bit unsigned
; integers and stores the sum in a variable
; Revision date:

INCLUDE Irvine32.inc

.data
source BYTE " This is the source string", 0 
target BYTE SIZEOF source DUP('#') 

.code
main PROC

 mov esi, (OFFSET source) + (SIZEOF source) - 2
 mov edi, OFFSET target
 mov ecx, SIZEOF source

L1: mov al, [esi]
 mov [edi], al
 dec esi              ; pointer to source string
 inc edi              ; pointer to target string
 loop L1
 
 mov esi, OFFSET target     ; offset of variable
 mov ebx, 1   ; byte format
 mov ecx, SIZEOF target-1 ; counter
 call Dumpmem
 
 exit

main ENDP
END main


This is the output that i get ::




the different is... there is extra space in the output compare the answer
that already give in the question.

to solve this... ** mov esi, (OFFSET source) + (SIZEOF source) - 2
                    change that 2 to 1.

                 ** mov esi, (OFFSET source) + (SIZEOF source) - (1)

then u'll get the output exactly like shown above in the question. :)




done. (",)v . got 10 marks for this.