Question

NOTE: explain the lines in comments for better understanding

Write an assembly program (for x86 processors - Irvine) that has two procedures, a main procedure and a procedure called Fib.

The fib procedure is to uses a loop to calculate and printout the first N Fibonacci numbers.

Fibonacci sequence is described by the following formula:

Fib(1) = 1, Fib(2) = 1, Fib(n) = Fib(n – 1) + Fib(n – 2).

The value of N is to be communicated to this Fib proceeddure via the EAX register

(i.e. main sets the value BEFORE it calls Fib) Fib should neatly prints out each

Fibonacci numbers, as it calculates them. It should put each number on its own line.

it should use WriteInt for the printing

(See http://programming.msjc.edu/asm/help/index.html?page=source%2Fabout.htm)

The main procedure need to ask the user HOW MANY of Fibonacci numbers to printout

and then reads in the value form the user (uses ReadDec -see bellow and maybe

http://programming.msjc.edu/asm/help/index.html?page=source%2Fabout.htm)

It should set the EAX register and then call the Fib procedure

Once the Fib procedure returns main should exit

.data

decNum DWORD ?

.code

read: call ReadDec

mov decNum,eax ;store good value

extra credit (optional)

We could do a better job of handling the input (invalid input)

Take a look at the documentation example from the above URL

and incorporate it your code

STARTER CODE FOR FIBONACCI SEQUENCE:

; Fib(1) = 1, Fib(2) = 1, Fib(n) = Fib(n – 1) + Fib(n – 2).

.data

prev DWORD 0

total DWORD 1

.code

Fib PROC USES EAX EBX ECX EDX

mov ecx, ; Loops N times

mov eax, ; puts prev into eax register (Fib(0) = 0 or n-2) into EAX

mov ebx, ; puts total into ebx register (Fib(1) = 1 or n-1) into EBX

L1:

mov ; move eax into edx register Fib(n) = Fib(n-2) EDX = EAX

add ; add prev and total Fib(n) = Fib(n-2) + Fib(n-1) EDX = EDX + EBX

mov ; prev = total save a copy for next iteration EAX = EBX

mov ; total = sum EBX = EDX

call WriteInt

loop L1

ret

Fib ENDP

Answer 1

public main

main proc

mov ebx, 0 ; first number
mov ecx, 1 ; second number
; edx will contain the third number (ebx + ecx )

mov eax, fib-2 ; print fib numbers (not counting the first
and second because they will be printed in the begining)

mov edx, ebx
call WriteInt ; printing the first number

mov edx, ecx
call WriteInt ; printing the second number
fib:
mov edx, ebx
add edx, ecx ; edx = eax + ebx

call WriteInt

; now we have the third number in edx
; ebx = 1st, ecx = 2nd, edx = 3rd
; to prepare ebx and ecx for the next iteration, shift the values to the right
; ebx = 2nd, ecx = 3rd, edx = ?
mov ebx, ecx
mov ecx, edx

loop fibo
ret

main endp

end main