问题
Today I wrote a recursive fibonacci in assembly and it doesn't work. I compiled it to object file with NASM and than made it elf with gcc.
When I enter 1 or 2 the function works perfectly, but when I enter 3, 4, 5, 6, or more the function doesn't work.
I think there is problem where the function calls itself.
This the code:
SECTION .data ;init data
str: db "This equal: %d",10,0
SECTION .text ;asm code
extern printf
global main
main:
push ebp
mov ebp,esp
;--------------------
push 03 ; the index
call _fibonacci
add esp,4
push DWORD eax
push str
call printf
;---------------------
mov esp,ebp
pop ebp
ret
This is the function:
_fibonacci:
push ebp
mov ebp,esp
mov ebx, [ebp+8] ;; param n
cmp ebx,0
jne CHECK2
mov eax,0
jmp _endFIbofunc
CHECK2:
cmp ebx,0x1
jne ELSE3
mov eax,1
jmp _endFIbofunc
ELSE3:
mov ebx,[ebp+8]
dec ebx ;; n-1
;; FIRST call
push ebx
call _fibonacci
add esp,4
mov edx,eax
;; SEC CALL
dec ebx
push ebx
call _fibonacci
add esp,4
add eax,edx
mov eax,[ebp-4]
_endFIbofunc:
mov esp,ebp
pop ebp
ret
After I ran it on Ubuntu 16.04 it send error:
Segmentation fault (core dumped)
What might be the problem?
回答1:
You must store (push) the registers you are going to change in the recursive call. And then restore their original values (pop). That should fix the problem.
Something like this:
- Push all registers you are going to use in your function (except eax where you will get return value)
- Push ebx as that is your parameter
- Call the function
- Add esp,4
- Pop all registers you pushed in the first step, now in the reverse order
回答2:
mov eax,[ebp-4]
You are using the memory at [ebp-4] without having put something useful in it!
You need to reserve this space in your function prolog:
_fibonacci:
push ebp
mov ebp, esp
sub esp, 4
Returning from the 1st recursive call you put the result from EAX in this memory dword.
Returning from the 2nd recursive call you add to EAX the contents of this memory dword.
Doing so, the EDX register will no longer be clobbered.
Why do you use the EBX register at all? If you use it you have to preserve it as was explained in the answer by Al Kepp.
If you start by putting the argument in EAX, you know that for both values below 2 (so 0 and 1), the result is just equal to the argument. Simple.
mov eax, [ebp+8] ;; param n
cmp eax, 2
jb _endFIbofunc
If you don't balance the stack directly after the 1st recursive call, you can just decrement the dword that is already there and make your second recursive call.
dec eax ; n-1
push eax ;(*)
call _fibonacci
mov [ebp-4], eax
dec dword ptr [esp] ; n-2
call _fibonacci
add esp,4 ;(*)
add eax, [ebp-4]
The whole proc:
_fibonacci:
push ebp
mov ebp, esp
sub esp, 4 ;(*)
mov eax, [ebp+8] ;; param n
cmp eax, 2
jb _endFIbofunc
dec eax ; n-1
push eax ;(*)
call _fibonacci
mov [ebp-4], eax
dec dword ptr [esp] ;(*) n-2
call _fibonacci
add esp,4 ;(*)
add eax, [ebp-4]
_endFIbofunc:
mov esp, ebp
pop ebp
ret
回答3:
In addition to the other answers provided, here's an alternate solution:
_fibonacci:
mov eax,[esp+4] ;eax = n
cmp eax,2 ;br if n < 2
jb _endFIbofunc
dec eax ;push n-1
push eax
call _fibonacci ;returns eax = fib(n-1)
xchg eax,[esp] ;eax = n-1, [esp] = fib(n-1)
dec eax ;push n-2
push eax
call _fibonacci ;returns eax = fib(n-2)
add eax,[esp+4] ;eax = fib(n-1)+fib(n-2)
add esp,8
_endFIbofunc:
ret
Trivia - fib(47) is the largest < 2^32. The number of recursive calls = 2*fib(n+1)-1.
n fib(n) # calls
0 0 1
1 1 1
2 1 3
3 2 5
4 3 9
5 5 15
6 8 25
7 13 41
8 21 67
9 34 109
10 55 177
11 89 287
12 144 465
13 233 753
14 377 1219
15 610 1973
16 987 3193
17 1597 5167
18 2584 8361
19 4181 13529
20 6765 21891
21 10946 35421
22 17711 57313
23 28657 92735
24 46368 150049
25 75025 242785
26 121393 392835
27 196418 635621
28 317811 1028457
29 514229 1664079
30 832040 2692537
31 1346269 4356617
32 2178309 7049155
33 3524578 11405773
34 5702887 18454929
35 9227465 29860703
36 14930352 48315633
37 24157817 78176337
38 39088169 126491971
39 63245986 204668309
40 102334155 331160281
41 165580141 535828591
42 267914296 866988873
43 433494437 1402817465
44 701408733 2269806339
45 1134903170 3672623805
46 1836311903 5942430145
47 2971215073 9615053951
48 4807526976 ...
来源：https://stackoverflow.com/questions/47001872/recursive-fibonacci-assembly