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: ' ;0 B, q) p" h5 L- Q3 { ~
lenUserMsg equ $-userMsg ;
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disp1Msg db 'The factorial of '
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lenDisp1Msg equ $-disp1Msg
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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.',0x0a3 W! H2 j6 ^4 M1 `8 P
lenErrMsg equ $-errMsg
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section .bss ;
num resb 1
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result resd 1
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buffer resb LEN_OF_OUTPUT+1" W/ _& f9 [, q9 A, W
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section .text ;+ K' B2 P' Q2 \- C9 P" h. w$ }
global _start
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_start: ;
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mov eax, 4
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mov ebx, 1& _% w' z, @- q* q' @# ?' X) b
mov ecx, userMsg4 q) u! z+ D# I: L) o& {8 V% D
mov edx, lenUserMsg
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int 80h$ i5 D, N* F2 _( U' U2 ^
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; read user's input/ }& E- _) X5 D u
mov eax, 3
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mov ebx, 0( P9 V) S6 e0 P( _+ g0 N
mov ecx, num ) S' {' Q5 T2 ~
mov edx, 1 ; length of input buffer
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int 80h4 g% Y+ b8 c. S' d- D
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mov eax, [num]) S( a! c, k# y) u) D9 ?6 d6 t& ?
and eax, 0xFF
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sub eax, 0x302 s6 R1 a- k% D* w% l3 ~
jl ERROR- `7 g0 f! s. n$ l& }
cmp eax, 10! R+ p1 O) F- q0 q
jl DO_CALC2 d1 N. P- n% }1 C
ERROR:' G& @; O9 c+ ]* m( t' }6 S3 }
mov eax, 4
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mov ebx, 12 {5 b! c+ E- ^. |" M% f
mov ecx, errMsg
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mov edx, lenErrMsg
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int 80h , g$ ^$ \! ] Y5 C
jmp EXIT
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DO_CALC:7 u0 P/ @4 U0 A0 K
call factorial
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; 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
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mov eax, 4
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mov ebx, 1
mov ecx, disp1Msg+ U$ m' u0 e# Y! p4 ?
mov edx, lenDisp1Msg
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int 80h
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mov eax, 46 @' k' I4 `/ _5 X) o- a- I4 h3 E
mov ebx, 1
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mov ecx, num
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mov edx, 1
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int 80h : o# W7 X) Y7 O0 k3 n" C
mov eax, 4
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mov ebx, 1
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mov ecx, disp2Msg
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mov edx, lenDisp2Msg4 u& @% ~% r+ L
int 80h 9 @& S6 ]# O& m6 E% T5 _' v( f
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; output number3 g2 a" H! ^8 ^% ?! F# a* ?
mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY( @1 D8 i. g9 B0 C' L: I7 }
mov byte [buffer+LEN_OF_OUTPUT],0x0a
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lea esi,[buffer+LEN_OF_OUTPUT]
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mov ebx,10 - S9 @1 `- c- e6 u
ASC_LOOP:
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mov edx,0 ; clear dx prior to dividing edx:eax by bx) G: _0 @4 b! f @- _0 { O9 x
div ebx ;DIV EAX/10
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add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER 6 \0 V& V* y! f$ ~" O
dec esi ; store characters in reverse order. J# Z5 m" L: i$ e6 B
mov [esi],dl" U! D1 C5 o2 ?, F
cmp eax,0 ; a) u) D; ^" y
jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII
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jmp ASC_LOOP ;ELSE REPEAT
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END_ASC:' o, l/ n/ c+ a3 y4 D% _
mov eax, 4
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mov ebx, 1
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mov ecx, esi/ p! U' u+ C' C1 q/ u
lea edx,[buffer+LEN_OF_OUTPUT+1]
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sub edx, ecx9 v A- ~& x2 X; [. ~3 u# j
int 80h : Y1 V3 r1 e6 [/ ?
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; Exit code( k8 ? H6 e9 `; n+ P+ n# M. b+ T8 k
EXIT:; {! m$ O1 t/ j; }2 \
mov eax, 1
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mov ebx, 02 q4 ]8 F3 o6 D- y% k8 m
int 80h, I s* [8 K, A3 g3 W* _) L: \$ V
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; Factorial function using recursion
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factorial:
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; Check if eax is 0 (base condition)( v/ d n. f9 W3 F) d8 F* p
cmp eax, 0
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jz end_recursion! l: Q" C! F7 R _
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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 recursively7 N( o+ |2 `6 Q# B
dec eax! v+ O% |" y! ?4 W; k
call factorial
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; Multiply the result returned in eax with the saved value of eax
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pop ebx0 p: Q) L9 l: C8 z* Q1 S
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)2 u. D. z4 Q6 m; j# X2 _0 u
mov eax, 1
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ret

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