用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 10
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section .data ;
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userMsg db 'Please enter a single-digit number: ' ;
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lenUserMsg equ $-userMsg ;
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disp1Msg db 'The factorial of '( F( S0 w; O3 K5 O' j
lenDisp1Msg equ $-disp1Msg / b5 [4 _- `9 E/ R
disp2Msg db ' is '; F" X/ G) z# x+ g6 e% c% R0 e
lenDisp2Msg equ $-disp2Msg @9 y6 [+ Z6 k. y1 R4 R0 q* ~ N
errMsg db 'Wrong input. Only one-digit number can be accepted.',0x0a
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lenErrMsg equ $-errMsg ! {% J7 R2 r6 R( L2 a5 l& P
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section .bss ;
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num resb 1! L- r; p3 }" z! i" b
result resd 15 @ S. w- ~( ?
buffer resb LEN_OF_OUTPUT+1- n X) j$ v: v9 a
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section .text ;
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global _start1 [& B/ _3 D& w
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_start: ;
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mov eax, 45 }& y: c" `! U( {/ F& ^- G7 Q! N
mov ebx, 1; g. r1 W- P/ ~4 C. {
mov ecx, userMsg
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mov edx, lenUserMsg5 ]! ]$ k/ F9 j! A+ B' r8 L. D; @
int 80h8 {# V7 z9 k7 w( H: t4 @. I
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; read user's input. ?5 _+ q2 v9 w% o/ Q6 v% T
mov eax, 3
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mov ebx, 01 A# Y- A3 K0 D* h, z; o& X: F
mov ecx, num ! J' O' h" |: Z, T9 A; j
mov edx, 1 ; length of input buffer
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int 80h
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mov eax, [num]
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and eax, 0xFF
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sub eax, 0x30
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jl ERROR/ |' u4 R& Z+ }& c; Y
cmp eax, 109 f4 _$ S# S6 `- ^$ ?8 a9 r u: [
jl DO_CALC
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ERROR:
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mov eax, 4
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mov ebx, 1
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mov ecx, errMsg6 d5 E7 k' V& R/ _
mov edx, lenErrMsg
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int 80h
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jmp EXIT
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DO_CALC:9 ~9 W8 e0 t$ D
call factorial* H6 E. {1 _0 c5 J/ }
; The parameter is in eax. After calling, the result will be stored in eax.
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mov [result], eax ; Save the result
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; print result1 a$ M9 L! ~6 ~, \4 r0 a. P
mov eax, 4
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mov ebx, 19 V! R- w8 O1 W' Z& G* ?$ R
mov ecx, disp1Msg
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mov edx, lenDisp1Msg
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int 80h
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mov eax, 4
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mov ebx, 1 u, w$ q0 }* P" s0 R, K) \
mov ecx, num
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mov edx, 1( R; _0 z% O3 _' O" L/ @2 n
int 80h
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mov eax, 4/ j. J/ U1 p/ y! U, d( ^
mov ebx, 1
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mov ecx, disp2Msg5 x5 S8 W I) Y* C8 U
mov edx, lenDisp2Msg1 v$ n: u+ B& Y! S
int 80h 3 Z; a* z' A6 z# j
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; output number3 t% H) K3 s! d, F O2 M) V
mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY$ U4 \9 u4 N. X$ p/ N6 ]4 K0 O
mov byte [buffer+LEN_OF_OUTPUT],0x0a P. a- @/ \: W- ~# s
lea esi,[buffer+LEN_OF_OUTPUT]3 ~* M1 o" g3 I, F
mov ebx,10 2 t4 h) ] J( B
ASC_LOOP:4 B9 T9 p8 N& N% N! }
mov edx,0 ; clear dx prior to dividing edx:eax by bx
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div ebx ;DIV EAX/10, L6 x% p2 l6 x! {3 h- s. J$ z
add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER . ]7 u* S4 @8 @ J, l& s
dec esi ; store characters in reverse order5 S' @: u, r, G+ r" C* E2 t+ ^" r
mov [esi],dl; v! }& _; }( k5 \5 z
cmp eax,0 , k8 ]$ p Q) K8 Q" U- ~
jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII
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jmp ASC_LOOP ;ELSE REPEAT" Z; H8 [9 J+ O# ?9 Q8 U. Q' c
END_ASC:3 [# h; R& \) \4 |) s& D# o
mov eax, 49 c9 v6 U3 J0 }1 y. k! g
mov ebx, 15 I' O) u( b4 ? |6 f
mov ecx, esi
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lea edx,[buffer+LEN_OF_OUTPUT+1]
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sub edx, ecx2 V/ K* I/ g7 N0 `6 ?: [
int 80h
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; Exit code
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EXIT:
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mov eax, 1' r% f* Y* k- F" Z. [! u
mov ebx, 06 ]2 R8 p* |$ a' S* h9 ]
int 80h
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; Factorial function using recursion( U# x# J) o% x
factorial:
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; Check if eax is 0 (base condition)) f9 T$ ]# D9 T6 U- S6 ~. l
cmp eax, 0
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jz end_recursion
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; Save the current value of eax
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push eax7 V |; b, Q) G
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; Decrement eax and call factorial function recursively0 [3 C4 \% [: B% G" f
dec eax) e) E# D2 V$ Y# u9 E' ~
call factorial% w C1 A1 W1 Q0 l+ i5 |. y
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; Multiply the result returned in eax with the saved value of eax
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pop ebx
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imul eax, ebx
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ret
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end_recursion:
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; Return 1 when eax is 0 (base condition)
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mov eax, 1, u g8 |( F H/ h+ c- G9 H
ret

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