用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 10- [( e: o3 h$ f' y# _
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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 '
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lenDisp1Msg equ $-disp1Msg - @* Z- e" Z* Y' Z
disp2Msg db ' is '
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lenDisp2Msg equ $-disp2Msg 0 B+ m2 L9 `9 Y0 w
errMsg db 'Wrong input. Only one-digit number can be accepted.',0x0a
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lenErrMsg equ $-errMsg
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section .bss ;
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num resb 1
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result resd 1' m# }5 R- J: H$ \* d) y& j% q, J! c
buffer resb LEN_OF_OUTPUT+1
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section .text ;
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global _start
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_start: ;, f) n6 g# o! z( t2 C( e
mov eax, 4 r0 [8 X s% z/ U& @' N
mov ebx, 1
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mov ecx, userMsg
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mov edx, lenUserMsg
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int 80h
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; read user's input" t5 j3 ]- ?5 m f
mov eax, 3
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mov ebx, 0
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mov ecx, num
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mov edx, 1 ; length of input buffer
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int 80h
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mov eax, [num]; H# U: j+ m! H5 y' j8 b4 n
and eax, 0xFF
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sub eax, 0x30
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jl ERROR R! O7 x9 F3 M& Y) e) j# B
cmp eax, 10( m5 y" @; t& d6 L, h0 G, R
jl DO_CALC
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ERROR:4 e( X( N% o4 M0 d7 h9 V& c
mov eax, 4
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mov ebx, 1; [3 `2 I. u& {. j- u
mov ecx, errMsg
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mov edx, lenErrMsg
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int 80h
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jmp EXIT
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DO_CALC:! R+ T( z0 m2 v8 v5 V; Y) w
call factorial n5 x3 J) I$ f
; The parameter is in eax. After calling, the result will be stored in eax.
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mov [result], eax ; Save the result/ t7 m3 |4 d0 E( q9 ^
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; print result
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mov eax, 4# y: O0 e* h# |$ P, W! n
mov ebx, 1! D9 \3 \2 F: }8 M9 A
mov ecx, disp1Msg f1 N% s1 e1 ?) \
mov edx, lenDisp1Msg& [3 t+ y- O; ]. ~1 B8 Z3 u6 i3 _ I+ j: R9 n
int 80h
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mov eax, 4$ a& ]" l7 W3 N, f, ~' H3 E
mov ebx, 15 Y6 p! t: Z# I, q" Y, J: l& x
mov ecx, num7 e: M4 k8 l4 }- i
mov edx, 1
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int 80h 0 k% k9 Y( h4 X& }/ A
mov eax, 4) \& R9 N S/ P2 n }: @
mov ebx, 1
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mov ecx, disp2Msg2 \6 H! p) V% |" L% Z0 y4 M
mov edx, lenDisp2Msg) z4 @" N7 U- |$ C+ l$ F/ I4 D
int 80h 2 J3 X/ C( S1 _7 y
; output number. W( j# [* \5 J. A3 \, E' t
mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY
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mov byte [buffer+LEN_OF_OUTPUT],0x0a
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lea esi,[buffer+LEN_OF_OUTPUT]% J' g( O; i- J5 X5 G
mov ebx,10 : l9 U. t7 W+ W9 V9 p* v& g
ASC_LOOP:7 q9 x9 n T+ i1 w: ^+ H6 |, X
mov edx,0 ; clear dx prior to dividing edx:eax by bx- @$ e+ C( q* t5 a- U: y
div ebx ;DIV EAX/10
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add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER 4 l1 u% `: f& f3 [8 R$ P/ N) u
dec esi ; store characters in reverse order: [/ q; j. B0 c' f2 ]8 u8 M
mov [esi],dl
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cmp eax,0
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jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII# ?( [* S- ~$ _$ B) z" V0 z
jmp ASC_LOOP ;ELSE REPEAT
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END_ASC:
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mov eax, 4 f4 k8 v# f. Q: t: y* ~" o1 f; C
mov ebx, 1. L. A+ w, g! M1 |" O! ?6 n7 x7 w
mov ecx, esi
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lea edx,[buffer+LEN_OF_OUTPUT+1]
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sub edx, ecx1 e3 ?+ n ~4 f
int 80h : v$ {3 Z& R* @0 r4 O) M
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; Exit code
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EXIT:- w" Z |: Z+ C, P3 n% W, I9 ^% r* ~
mov eax, 1
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mov ebx, 0
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int 80h
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; Factorial function using recursion1 F) n9 y! K! }" r6 J8 ]5 e
factorial:
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; Check if eax is 0 (base condition)4 ?0 Q1 ~+ g+ f: m
cmp eax, 06 V* u, V" Q' M
jz end_recursion
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; Save the current value of eax: a2 V4 [2 R9 ]& j2 C5 @
push eax
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; Decrement eax and call factorial function recursively0 G9 Y" B7 ?2 G
dec eax$ A7 r: `2 ^& j* n- K; x& y8 _# f
call factorial
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; Multiply the result returned in eax with the saved value of eax$ G% X8 B% m- Y" T4 u
pop ebx
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imul eax, ebx. i3 [ |3 |; P9 c3 u3 G% r4 G
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ret6 z3 y3 v- P$ A9 [) g$ d
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end_recursion:7 a' l5 B5 S! P6 j' g" S
; Return 1 when eax is 0 (base condition)5 h. [+ m0 }9 x3 g$ u
mov eax, 1
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ret

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