【发布时间】:2016-06-05 10:14:59
【问题描述】:
经过漫长的无数次错误,希望这是最后一个。
没有编译或运行时错误,只是逻辑错误。
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
这将打印系列的最终结果。
我的汇编代码:
我尽可能添加评论
.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,但输出不让我满意:
输出:Loop Ran : 10 Times -----Final Number is : 11 ------
首先我不确定最终数字是十进制还是十六进制。
- 假设为十进制:斐波那契数列没有 11
- 假设它是十六进制:斐波那契数列没有 17(十六进制 11 = 十二月 17)
我做错了什么?
【问题讨论】:
-
无需不确定打印的数字是否为十进制。
printf使用finalprint字符串作为格式,如果它类似于常规的printf,它将使用%d以十进制形式输出。 -
只需将您的 cmets 与您真正想做的比较;)
NOW second=first是的,但您想要first=second... 哎呀。您可以通过评论获得 +1,这就是我们发现您的错误的方式。 -
注意:伪代码返回正确的斐波那契数,尽管对于 n=10 它返回
144,技术上是 12th fib num(或89,取决于如何n已初始化,但仍差一点)。 -
@Jester 谢谢,我会记住这一点,下次:)
-
@RadLexus 感谢您的信息:)