在一些作业中,我必须在装配中创建一个斐波纳契序列程序。我创建了这个代码,但它似乎没有正常工作,我不确定为什么。我相信我正在正确这样做,但EAX仍然是每圈“2”。

    INCLUDE Irvine32.inc
    .data
        prev DWORD ?
        next DWORD ?
        val DWORD ?
        count DWORD ?
        total DWORD ?

        myMsg BYTE "Fibonacci Sequence ",0dh,0ah,0

   .code
    main PROC
       mov ecx,15
       mov val,1
       mov prev,-1
       mov eax,1
       mov edx,OFFSET myMsg
       call WriteString

    L1:
       mov count,ecx
       mov ebx,val
       add ebx,prev
       mov total,ebx
       mov ebx,val
       mov prev,ebx
       mov eax,total
       mov val, ebx
       call WriteInt
       call Crlf
       loop L1

    exit
    main ENDP
    END main
.

有帮助吗?

解决方案

看起来像这样(未经测试):

    mov  ecx, 15
    mov  eax, 0    ;a = 0
    mov  ebx, 1    ;b = 1
_fib:
    mov  edx, eax 
    add  edx, ebx  ;sum = a + b
    mov  eax, ebx  ;a = b
    mov  ebx, edx  ;b = sum
    loop _fib
.

其他提示

您的循环在伪代码中简化为此:

L1:
   count = ecx; // count === 15
   eax = total = val + prev; // prev = -1 => eax = 0. prev = 1 => eax = 2
   prev = val; // sets prev = 1, val doesn't change so prev = 1 after the first iteration
.

如您所见,EAX= VAL + PREV将评估为2,一旦PREV设置为1。

你应该详细说明你的问题的规范。你想打印多少个整数?这是什么= 15是为了?在这种情况下,您需要在每次迭代和检查时都需要减少计数,并检查它是非零。

对于fibonacci序列,您应该在循环中做一些这样的事情:

// lets say that eax is the current integer in the sequence and prev is the previous integer
// then the next integer = eax + prev
ebx = eax + prev
prev = eax
eax = ebx
.

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