Question

我试图在x86 NASM程序集中实现递归Ackermann-Peter函数。函数定义如下：

*a（0；m）=m+1

*a（n+1；0）=a（n；1）

*a（n+1；m+1））=a（n；a（n+1；m））

我的问题是我甚至无法想象如何正确地开始。到目前为止，我只在Assembly中递归地实现了一个“x的幂”函数。

Here is what i have so far: http://pastebin.com/rsWALyCq (The german prompts just ask for n and m)

我很感激我能得到的每一点帮助。

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所以我现在做了push/pop Statements Symetric，但仍然存在Segmentation错误。我试着调试整件事，并在第一个案例中放置了一条调试消息。我编译了这个程序，并在n=0和m=0的情况下进行了尝试，但他没有打印调试消息，所以他甚至没有输入第一种情况。我似乎弄不明白他为什么不这么做。

Heres my current try: http://pastebin.com/D4jg7JGV

Answer 1

解决方案：

好的，我发现了问题：

I didn t manage the ebp and esp right. So, I now used the ENTER and LEAVE Macros in every function call and now the whole thing works properly. Here is the Solution. Thank you for your time:

asm_main:
    ENTER 0,0               ;setup Routine
    PUSHA
    MOV eax, prompt1        ;ask for n

Full Code: http://pastebin.com/ZpPucpcs

Answer 2

如果您可以递归地执行此操作（伴随着所有堆栈帧的增长），那么这就相当简单了。子程序代码中的基本思想是：

ACKERMANN
Get n and m off the stack or from registers
Compare n to zero.
If it is equal, jump to FIRSTCASE
Compare m to zero
If it is equal, jump to SECONDCASE
put n + 1 into the first argument slot
put m into the second argument slot
call ACKERMANN
put n into the first argument slot
put the return of previous call into second argument slot
call ACKERMANN
put the return of the previous call into the return register/stack slot
return

FIRSTCASE
put m+1 in the return register/stack slot
return

SECONDCASE
Put n into the first argument slot
put 1 into the second argument slot
call ACKERMANN
put the return of previous call into return register/stack slot
return

友情链接