Fibonacci Series in Assembly x86

Question

Finally after a long session of countless errors , hope this is the final one.

No Compile or runtime errors, Just a logical error.

EDIT: (Fixed Pseudocode)

Answer 1

The problem was that my Actual code was not matching with my Pseudocode which resulted in the Logical error.

This Part

     mov ebx,first       ;move first into ebx
     mov second,ebx      ;copy ebx to second [NOW second=first]

This gives first value of second, but my PseudoCode says "first=second", which means give value of second to first.

     mov ebx,second      ;move second into ebx
     mov first,ebx       ;copy ebx to second [NOW first=second]

Final Working Code for x86 Intel Processor:

For any further referrers , i am posting a working code for x86 intel

.386
.model flat,stdcall
option casemap:none

.data
timestell   db   "Loop Ran : %d Times -----",0          ;format string
finalprint  db   "%d th Fibonacci number is %d",0       ;format string
times       dd   14h                                    ;times to loop
first dd 1h
second dd 1h
third dd 0h



.code
include windows.inc
include user32.inc
includelib user32.lib
include kernel32.inc
includelib kernel32.lib
includelib MSVCRT
extrn printf:near
extrn exit:near

public main
main proc


         mov ecx, times       ;set loop counter to "times" time
         sub ecx,2            ;loop times-2 times

      top:
         cmp ecx, 0          ; test at top of loop
         je bottom           ; loop exit when while condition false
         xor ebx,ebx         ;Clear ebx
         mov ebx,first       ;move first into ebx
         add ebx,second      ;add ebx, [ first+second ]
         mov third,ebx       ;Copy result i.e ebx [first+second] to third
         xor ebx,ebx         ;clear for further use
         mov ebx,second      ;move second into ebx
         mov first,ebx       ;copy ebx to second [NOW first=second]
         xor ebx,ebx         ;clear for later use
         mov ebx,third       ;move thirs into ebx
         mov second,ebx      ;copy ebx to third [NOW second=third]
         xor ebx,ebx         ;clear it
         dec ecx             ;decrement loop
         jmp top             ;Loop again

      bottom:
        push third
              push times                ;push value of third to printf
              push offset finalprint    ;push the format string
              call printf               ;Print the final number
      push 0        ;exit gracefully
         call exit      ;exit system

    main endp

end main