组装中的非递归斐波纳契序列
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09-12-2019 - |
题
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
. 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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