装配x86中的斐波纳契系列

最后在经历了无数次的错误之后,希望这是最后一次。

没有编译或运行时错误,只是一个逻辑错误。

编辑:(固定伪代码)

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。

  • 假设它是十进制的:斐波那契数列没有11
  • 假设它为十六进制:Fibonacci Series没有17(11十六进制= 17十进制)
  • 我究竟做错了什么?


    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]
    

    这给了second first值,但我的PseudoCode说“first = second”,这意味着给secondfirst值。

         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,我发布了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
    
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