Question

我试图在程序集上实现下面的函数作为递归函数调用。

Int  f(n) 
if (n<=3)   return n;
else        return  2 * f(n-1) + f(n-2);

当我运行代码时，我收到1作为结果。请建议

INCLUDE Irvine32.inc
.CODE
main PROC
    push 5
    call RECURSIVE          ; calculate the below function
    call WriteDec           
    call Crlf
    exit
main ENDP
;------------------------------
RECURSIVE PROC
; Calculates int f(n)
; if (n<=3) returns n
; else return 2*f(n-1)+f(n-2)
;------------------------------
push ebp                    
mov ebp,esp             
mov eax,[ebp+8]             ;get n
cmp eax,3                   ;n > 3 ?
ja L1                       ; yes, continue
mov eax,1                   ; no, return 1 
jmp L2                      ;return to the caller
L1:  dec eax
     push eax
     call RECURSIVE
ReturnFact:mov ebx,[ebp+8]  ;get n-1
           shl ebx,1        ;multiply by 2
           add ebx,[ebp+16] ;add n-2 and save 

L2: pop ebp                 ;return eax
    ret 4                   ;clear stack
RECURSIVE ENDP

END main

Answer 1

INCLUDE Irvine32.inc
.CODE
main PROC
    mov ecx, 7
    push ecx
    call RECURSIVE      
    call WriteDec           
    call Crlf
    exit
main ENDP
;------------------------------
RECURSIVE PROC
; Calculates int f(n)
; if (n<=3) returns n
; else return 2*f(n-1)+f(n-2)
;------------------------------
push ebp                    
mov ebp,esp             
mov eax,[ebp+8]             ;get n
cmp eax,3                   ;n > 3 ?
ja L1                       ; yes, continue
mov eax,[ebp+8]             ; no, return 1 
jmp L2                      ;return to the caller
L1:  dec eax
     push eax
     call RECURSIVE

ReturnFact: shl eax,1
            push eax 
            mov eax, [ebp+8]
            sub eax, 2       
            push eax 
            call RECURSIVE 
            add eax, [ebp-4] 
L2: mov esp, ebp
    pop ebp                 ;return eax
    ret 4                   ;clear stack
RECURSIVE ENDP

END main

汇编语言计算递归函数

1 个答案: