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 $-userMsg ;3 a$ k: P+ l* N8 x& R# I' i# z
disp1Msg db 'The factorial of '
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lenDisp1Msg equ $-disp1Msg
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disp2Msg db ' is '+ j: U: G' o7 ?1 M% ~- a
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 $-errMsg % H8 |, C( N+ a" ]+ [# k
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section .bss ;8 N" R$ n! n% {1 P% y& A; Z5 @
num resb 1. ^# M3 W9 A5 e: L) i
result resd 1
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buffer resb LEN_OF_OUTPUT+1
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section .text ;
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global _start# n7 g# J( U3 ?7 i' b. J, g
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_start: ;
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mov eax, 4! \7 m8 M' U7 v! m; G
mov ebx, 14 \, s) v9 |. ?& e Z$ j8 N" s- b
mov ecx, userMsg! {& }) d- M0 t F* `
mov edx, lenUserMsg* g/ n( ?1 B& s' Z+ x
int 80h [; c7 y& l7 k' m
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; read user's input X: {, `$ G5 K
mov eax, 3) ], b2 e/ h& m& s. f+ j
mov ebx, 0- A- _' z- d, ^% {. {& |
mov ecx, num 7 F' m' I( S! g5 Q
mov edx, 1 ; length of input buffer
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int 80h0 r f* V3 J. b' H$ E
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mov eax, [num]; G& p( d% @% V! v; `$ p
and eax, 0xFF
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sub eax, 0x30
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jl ERROR+ f9 e6 E5 L9 S* }* t
cmp eax, 10
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jl DO_CALC
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mov eax, 4" U7 I$ n) |0 p; f5 v, B
mov ebx, 1
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mov ecx, errMsg9 ]' @' r6 {2 S7 o
mov edx, lenErrMsg8 a7 }, O/ w2 |, c8 }8 M; ?
int 80h
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jmp EXIT
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DO_CALC:. u1 m& [) _# m, i- t, k1 i
call factorial: S0 l+ K4 a+ {2 h: e- `
; The parameter is in eax. After calling, the result will be stored in eax.+ T/ S+ p& b3 i% V) ?% v
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/ K l5 |# \& c) g4 p& g# C/ F
mov ecx, disp1Msg' s; {& w% o9 l& O4 I. `
mov edx, lenDisp1Msg, l% K$ U" u3 e" }! \0 \
int 80h * d4 S- P1 S1 U
mov eax, 43 Y D T; k" _7 M1 L0 t5 g
mov ebx, 1
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mov ecx, num) g" ]& e: I- J% A) r+ ]: |
mov edx, 1
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int 80h & Q) v& s6 z, r' S k* m1 D& i! X0 D+ X
mov eax, 49 s; ] s, W( Q* d. G
mov ebx, 1- F" m2 T/ V8 c8 S& P% w
mov ecx, disp2Msg
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mov edx, lenDisp2Msg& D9 y8 q0 ?" {- i. @
int 80h 4 A7 O r9 h/ ?7 D1 K) ]5 u
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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- ^* \& L1 ] H" t
lea esi,[buffer+LEN_OF_OUTPUT]
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mov ebx,10 / F# e$ _2 k6 O' c1 ^. O/ ~# N
ASC_LOOP:" N1 }" D. S2 p, g0 F
mov edx,0 ; clear dx prior to dividing edx:eax by bx
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div ebx ;DIV EAX/10, w, k( Y$ m5 g' y0 Q
add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER 0 _3 q3 l( Z0 m! w4 e
dec esi ; store characters in reverse order
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mov [esi],dl3 o- k/ n" a# n6 Q
cmp eax,0 ; h8 r/ K. M; j0 K& p K6 o
jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII+ K0 I6 c5 L8 T& f6 Q% L6 h3 k
jmp ASC_LOOP ;ELSE REPEAT, D8 I+ a! B. C. h2 g
END_ASC:
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mov eax, 4
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mov ebx, 16 G% ~$ _. B* b3 p0 ]5 `
mov ecx, esi
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lea edx,[buffer+LEN_OF_OUTPUT+1]
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sub edx, ecx ]: p( E: J; z" t* a
int 80h 3 J* O! g4 R- `% z2 e' j
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; Exit code& f& d8 e5 t- M% a& z
EXIT:3 E% h% _$ D3 E
mov eax, 1
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mov ebx, 0
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int 80h9 a9 j: S6 t( F- q. r
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; Factorial function using recursion
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factorial:. P" z8 N& r0 w1 t4 l2 u
; Check if eax is 0 (base condition)9 _- {) s' k* T3 ?5 O" T0 ~
cmp eax, 00 T! {, O9 b! w- M& f, X3 W
jz end_recursion! N; Z& @2 p' u7 o
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; Save the current value of eax( [; q2 j w( \0 N2 h+ J4 z
push eax
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; Decrement eax and call factorial function recursively
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dec eax) q9 ^. U) S/ x) j9 Y+ H
call factorial5 l7 w+ L: J& m+ a
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; Multiply the result returned in eax with the saved value of eax
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pop ebx: w0 u* f8 C2 V' g$ z- H1 V1 x
imul eax, ebx3 X( }+ A& n4 N( F+ e
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ret
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end_recursion:) Z( f, ~9 o2 C$ s1 \& t4 u
; Return 1 when eax is 0 (base condition)# c0 s) L# Q( U3 r N H
mov eax, 10 g' D7 A; J N+ E( a3 U2 p$ L
ret

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