用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 10
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section .data ;# c5 G% o$ c# M" y# I
userMsg db 'Please enter a single-digit number: ' ;& y9 t# y6 Y0 I" ]
lenUserMsg equ $-userMsg ;* M }6 y' p) ]( p: |; |
disp1Msg db 'The factorial of '
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lenDisp1Msg equ $-disp1Msg ) _5 B- x: @. b# m+ X9 a% i
disp2Msg db ' is '
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lenDisp2Msg equ $-disp2Msg
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errMsg db 'Wrong input. Only one-digit number can be accepted.',0x0a
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lenErrMsg equ $-errMsg & g" s) _6 E+ d
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section .bss ;
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num resb 1
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result resd 1
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buffer resb LEN_OF_OUTPUT+1$ a$ f2 ~9 U6 Y) i6 P* v- B
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section .text ;
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global _start
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_start: ;
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mov eax, 4$ q4 ], H( J, c G$ p$ p- r0 ]* I
mov ebx, 18 x# Z9 ^3 M5 S& O- L) {& L9 C
mov ecx, userMsg" ~4 H% E1 O2 a1 K3 ^+ y3 }
mov edx, lenUserMsg
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int 80h
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; read user's input; Z. p: \; p. C7 `8 z" B0 c
mov eax, 3
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mov ebx, 09 S3 \# G8 N5 Z5 D$ j
mov ecx, num
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mov edx, 1 ; length of input buffer7 `- N1 v- `/ b% ]! Z
int 80h
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mov eax, [num]' b4 ]1 r0 e7 `
and eax, 0xFF" G. d- u* A6 Y: v* {
sub eax, 0x30
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jl ERROR
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cmp eax, 10" y5 P2 ^% ?2 D/ y8 @. U$ T* A
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, errMsg
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mov edx, lenErrMsg7 F9 N2 `# m9 v- w7 x3 ]7 H
int 80h 6 j: [) q, W8 G% f
jmp EXIT
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DO_CALC:6 e( a% H& ^! P W1 m$ K
call factorial7 a( ]/ }7 @. l" s+ V0 n8 t+ F; m
; 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 result( Y. ]8 |: r n( L6 s
mov eax, 4% d8 G9 j0 T. ?/ N) o g1 P- _
mov ebx, 1( c: T. \' p4 ]& r1 l
mov ecx, disp1Msg8 O+ z) q- R8 m/ R( @8 {6 M6 Q* U
mov edx, lenDisp1Msg
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int 80h / |/ q2 `9 F) P, k- I& V4 f/ k
mov eax, 4/ y! X& a5 f6 Z# V K5 A. e
mov ebx, 1
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mov ecx, num
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mov edx, 1& C$ g7 g8 `/ C. U. P
int 80h . X) r: J" [5 E3 W6 z
mov eax, 4# i5 a+ L; l1 a( ]# |- h* m% p
mov ebx, 1
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mov ecx, disp2Msg" g' o: v8 S& f- A2 S/ {
mov edx, lenDisp2Msg/ U8 |3 K" d2 b: _
int 80h
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; output number
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mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY
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mov byte [buffer+LEN_OF_OUTPUT],0x0a$ C2 j' P- L& \3 v3 |: e! j. n
lea esi,[buffer+LEN_OF_OUTPUT]% A; ?; B% B6 d1 i f1 p
mov ebx,10
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ASC_LOOP:
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mov edx,0 ; clear dx prior to dividing edx:eax by bx, K+ `# s5 c, j/ \% Z3 e
div ebx ;DIV EAX/10
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add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER
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dec esi ; store characters in reverse order
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mov [esi],dl7 I& Q2 T Z D+ S+ `" Q
cmp eax,0 ( E" A+ w+ k; ?. e }* G$ ]
jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII' @' j& ~) Y* m1 f; ?. f! N0 d
jmp ASC_LOOP ;ELSE REPEAT! ?4 L9 K" p3 j$ ]9 Z4 f) F
END_ASC:! _& a/ X6 N5 G! y: g. p% ^& S
mov eax, 4
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mov ebx, 1, E1 x4 K3 j; d, G
mov ecx, esi# x% \" A! Y# H$ V r6 c
lea edx,[buffer+LEN_OF_OUTPUT+1]
sub edx, ecx
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int 80h - P$ s! b& h% ] ^* i: Q
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; Exit code
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EXIT:5 G9 L0 m) X, n- F Z* {6 m
mov eax, 1
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mov ebx, 0
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int 80h# c% ]/ \" \0 m' L" M$ c& S" B. m
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; Factorial function using recursion {/ {7 I' J8 y3 C5 d
factorial:
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; Check if eax is 0 (base condition)9 `9 g7 b* \* s: V1 n- `
cmp eax, 01 `, M( `4 u3 [2 x7 X$ r
jz end_recursion
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; Save the current value of eax
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push eax
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; Decrement eax and call factorial function recursively1 r( @- }: G: \* [9 }
dec eax8 ~6 _0 [. k$ r' [$ i9 B" _
call factorial
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; Multiply the result returned in eax with the saved value of eax
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pop ebx, l& f ^( {: h8 ^) S4 Q% R
imul eax, ebx
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ret/ d7 z/ o+ j' e7 z* X
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end_recursion:7 i& d# ~& t2 P) H5 T
; Return 1 when eax is 0 (base condition)
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mov eax, 15 j# |, L9 Q1 H3 r; i
ret

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