装配x86中的斐波那契系列

Question

经过长时间的无数错误之后,希望这是最后的错误.

没有编译或运行时错误,只是一个逻辑错误.

编辑:(固定伪码)

我的伪代码:

first  = 1;
second = 1;
third  = 0;

 for i from 1 to n{

    third=first+second
    first=second
    second=third

}
return third

这将打印该系列的最终结果.

我的汇编代码:

我尽可能地添加了评论

.386
.model flat,stdcall
option casemap:none

.data
timestell     db "Loop Ran : %d Times -----",0     ;format string
fmtd   db "%d",0
finalprint  db "Final Number is : %d ------",0     ;format string
times  dd 0Ah                                      ;times to loop
first dd 1h
second dd 1h
third dd 0h


.data?

retvalue1 dd ?             ;we will initialize it later

.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      ;loop "times" times
         mov eax,0           ;just to store number of times loop ran
      top:                   ;body of loop
         cmp ecx, 0          ;test at top of loop
         je bottom           ;loop exit when while condition false
         add eax,1           ;Just to test number of times loop ran
         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,first       ;move first into ebx
         mov second,ebx      ;copy ebx to second [NOW second=first]
         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:
           mov retvalue1,eax       ;store eax into a variable
           push retvalue1          ;pass this variable to printf
           push offset timestell   ;pass Format string to printf    
           call printf             ;Print no.  of times loop ran
           push third              ;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

代码运行良好,但输出不满足我:

输出: Loop Ran : 10 Times -----Final Number is : 11 ------

首先,我不确定最终号码是十进制还是十六进制形式.

• 假设它是十进制:斐波那契系列没有11
• 假设它是十六进制:Fibonacci系列没有17(11 hex = 17 dec)

我究竟做错了什么？

Answer 1

问题是我的实际代码与我的伪代码不匹配，导致了逻辑错误。

这部分

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

这给出了 的first值second，但我的伪代码说“first=second”，这意味着给出值secondto first。

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

x86 Intel 处理器的最终工作代码：

对于任何进一步的推荐人，我正在发布 x86 intel 的工作代码

.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