用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 105 Z/ c) p9 P6 [4 I' _" G
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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 ;8 [& _( }. \: ]% e+ L1 D
disp1Msg db 'The factorial of '9 ^3 z4 Q9 l# @
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.',0x0a! A3 e/ v. z1 B& x: v
lenErrMsg equ $-errMsg / J# @- ~0 z7 g
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section .bss ;
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num resb 1) o4 t' i. R7 q; ]2 u3 j. I+ W: R
result resd 17 f& D1 J4 k/ E; }5 a
buffer resb LEN_OF_OUTPUT+1
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section .text ;% O) N, @) w+ j1 a9 q! o$ j
global _start0 t; P' P2 c! @" u, p; M: M: p5 d
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_start: ;+ @. \1 p3 _: o$ O: o' h
mov eax, 4
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mov ebx, 1/ n' w! |. o* A: H9 }) t
mov ecx, userMsg
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mov edx, lenUserMsg
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int 80h& G% D( u& t1 ?6 r0 `
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; read user's input
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mov eax, 35 q" t- N& u* i- U5 T
mov ebx, 0
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mov ecx, num - J7 d% Z, B+ n* O9 n" C/ r
mov edx, 1 ; length of input buffer+ L# T4 ^! O1 B7 t8 G" v& s. Y3 I
int 80h8 O5 C# N( Y( J4 m# I
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mov eax, [num]' l% p' \6 A; R% k" C: J& a
and eax, 0xFF% O* _, ]6 L0 K4 L! m8 v
sub eax, 0x30
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jl ERROR+ m$ ` T* h1 l9 p
cmp eax, 10( Y/ ?$ p' Y! J0 l+ T& @
jl DO_CALC
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ERROR:+ k* i" ^- A: H* P7 p' ?
mov eax, 40 S* r5 N7 k8 L' ^+ F( X, B# S, c
mov ebx, 1/ ^! P/ f5 ?; l, e
mov ecx, errMsg: X5 z4 G2 }& Y+ p6 P2 ^
mov edx, lenErrMsg3 i y+ W6 f D
int 80h : D- I# V: `: g* V* j0 z+ e, Y
jmp EXIT& o4 F( Q- h/ r* p5 T! C
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DO_CALC:
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call factorial) p9 \7 K1 x) B) A3 l6 l3 S) q! W( u1 i
; 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) Z ?% S, y: ]9 s
mov ecx, disp1Msg6 f+ G0 k: c4 J }
mov edx, lenDisp1Msg$ r8 a' J3 \( J$ q, s
int 80h
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mov eax, 4! a- z6 }1 F: D- l6 H, \
mov ebx, 1
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mov ecx, num
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mov edx, 1
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int 80h + J) j) ^4 r, V$ C
mov eax, 44 i0 a8 y* k+ Z. B; A8 n
mov ebx, 1 i% K* s: }3 }5 Y. G& w
mov ecx, disp2Msg; T& c! I4 I# k# _7 {+ R8 J3 R
mov edx, lenDisp2Msg
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int 80h
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; output number
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mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY" T. ?+ X3 w7 A4 Q7 D
mov byte [buffer+LEN_OF_OUTPUT],0x0a6 r2 U! O/ V& b
lea esi,[buffer+LEN_OF_OUTPUT]/ U. W5 O7 d D6 `9 E4 F
mov ebx,10 % ?3 g# [- x) s4 \9 q+ _4 ~& a- ~4 I
ASC_LOOP:
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mov edx,0 ; clear dx prior to dividing edx:eax by bx
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div ebx ;DIV EAX/10! y% w+ G5 Y: [ ?$ g
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],dl9 b' j+ c1 Y! v# K- s+ r, \
cmp eax,0
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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:+ C7 Z* s& I6 _ g9 e
mov eax, 48 ^; k7 [) J4 F) P; m1 p Z" e
mov ebx, 1
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mov ecx, esi
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lea edx,[buffer+LEN_OF_OUTPUT+1]" I4 @" r' l) Q& |
sub edx, ecx
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int 80h 1 \ A$ w i7 I) ]
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; Exit code, F) ~ W: d9 b$ R
EXIT:
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mov eax, 1# B5 _5 t3 n5 L9 ~) V) D+ x1 H4 u
mov ebx, 01 q, B! Y. o" V+ X4 t
int 80h0 `# L W# t4 L
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; Factorial function using recursion( |4 p1 b' u# N7 u* O" z& z
factorial:( Z" \4 ~" P/ v* S) ?5 ?# C: Z
; Check if eax is 0 (base condition)& ^0 r2 p$ S7 h' J
cmp eax, 0
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jz end_recursion0 p1 G" ^( t" Z: B/ a6 _
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; Save the current value of eax. I5 m4 |2 l1 e
push eax" G- {. M0 b, R8 H# x) w, N8 d! }
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; Decrement eax and call factorial function recursively) l# ^8 q1 W: h' y
dec eax
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call factorial
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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)* {5 e1 d; N: {- ]; w
mov eax, 1
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ret

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