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: ' ;" W% b0 x3 v7 M1 o
lenUserMsg equ $-userMsg ;2 H' m. S, I5 C, f7 q7 ^
disp1Msg db 'The factorial of '
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lenDisp1Msg equ $-disp1Msg
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disp2Msg db ' is '# t H- o* k9 m$ @
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 $-errMsg # i4 N* o! z5 f/ L1 o3 L3 }" A" O
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section .bss ;/ i @& u" r: L5 y8 m2 n- V
num resb 1
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result resd 1. D5 K' N- M' N7 p8 R
buffer resb LEN_OF_OUTPUT+15 T- U5 c* b+ J& g4 { d
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section .text ;
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global _start
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_start: ;
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mov eax, 4
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mov ebx, 16 R" j; w6 p4 k. Z1 e7 }5 T3 L$ p
mov ecx, userMsg$ a8 ^9 ^4 P9 V. \5 z' ^' M3 b
mov edx, lenUserMsg
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int 80h
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; read user's input
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mov eax, 33 s. K) C4 E# V9 N7 y2 L2 ~$ h
mov ebx, 0# k Z3 f0 U9 h8 i1 w+ ]& E
mov ecx, num
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mov edx, 1 ; length of input buffer8 }5 T% I- C2 P- c0 k0 K3 Z1 c$ o
int 80h" t# Q3 T" m: o* b3 Z
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mov eax, [num]- M2 t9 U0 R& K( N9 e" o! { v
and eax, 0xFF: N7 ~( r8 q& Q% S- o: N4 P6 w
sub eax, 0x301 L6 \ ^$ \* w# C. V
jl ERROR K" \( l( C) z/ |+ Z# r, e5 k" X, w
cmp eax, 10
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jl DO_CALC: ^" ?+ w5 a& b; l/ n$ S6 Q2 u
ERROR:, S; k6 {/ ]! Y. {% b
mov eax, 4
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mov ebx, 1" f7 i% z+ L: u& N% u+ l/ }
mov ecx, errMsg1 l4 \# C) K6 Q4 |1 z
mov edx, lenErrMsg
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int 80h 5 P7 f6 {! Q# t* o& x
jmp EXIT$ i0 T% L _8 O: e3 @
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DO_CALC:6 O+ X" R; T1 ^; y
call factorial1 Y5 S+ l; y2 ]; i2 [
; The parameter is in eax. After calling, the result will be stored in eax.
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mov [result], eax ; Save the result' [. w. |% O9 x7 q8 N8 [* o
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; print result
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mov eax, 45 f9 }$ o1 i8 I
mov ebx, 1
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mov ecx, disp1Msg
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mov edx, lenDisp1Msg
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int 80h
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mov eax, 42 d1 j& I% ]8 |% R8 w
mov ebx, 1* k7 r( r q% }6 ~5 L9 I* G
mov ecx, num/ |* r' y/ L# @1 R) J9 c/ Q
mov edx, 1
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int 80h
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mov eax, 4
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mov ebx, 1
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mov ecx, disp2Msg
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mov edx, lenDisp2Msg/ I! }# F0 x" o6 f; t' e/ q+ P
int 80h
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; output number# n1 c) Z, x$ @" P
mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY
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mov byte [buffer+LEN_OF_OUTPUT],0x0a y" n4 p' N7 @( m; ?# }
lea esi,[buffer+LEN_OF_OUTPUT]
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mov ebx,10 9 n# l6 L2 Q3 W* G7 p
ASC_LOOP:8 Q) N9 s9 ^! ?% T7 x$ v/ K+ i
mov edx,0 ; clear dx prior to dividing edx:eax by bx
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div ebx ;DIV EAX/10
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add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER ( s/ h3 \. d4 o& V
dec esi ; store characters in reverse order
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mov [esi],dl& \, K; p" [, v1 t, Y& Q8 P& _
cmp eax,0 $ M& f$ l5 H' g* e, l# D. W
jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII; Z$ i1 W3 h" _0 E& \# O; G/ u3 K
jmp ASC_LOOP ;ELSE REPEAT
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END_ASC:/ G6 [/ ]# ^" T4 [' ?+ y2 h$ e
mov eax, 4% I3 p5 l6 t' Q& z, I0 t/ P
mov ebx, 1, \! Z g) S/ y$ w0 A
mov ecx, esi( m7 S) k9 f3 I5 M7 g" j$ P
lea edx,[buffer+LEN_OF_OUTPUT+1]7 O7 Z; j% j, w; J; z2 e
sub edx, ecx6 c; M. C# ^' D. S
int 80h 0 q6 y8 \, X% D7 n
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; Exit code
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EXIT:
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mov eax, 13 h( _* b' c: z4 [9 K! K
mov ebx, 0
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int 80h( h' L. ` ]7 t: S9 W
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; Factorial function using recursion7 v2 k: k B/ `7 R. j
factorial:- k! m, _; `. \5 e+ j7 h/ P& F+ R
; Check if eax is 0 (base condition)- y- i$ f+ L6 y' N+ Q! _# W6 a3 {. W
cmp eax, 0
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jz end_recursion8 \/ \! j; ^/ r8 D' D8 w
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; Save the current value of eax) D- E% o Z# X$ A
push eax' @3 H3 K4 M& B4 Q+ t5 v
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; Decrement eax and call factorial function recursively5 B# n- G: n: Z- L$ ?( r
dec eax
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call factorial
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; Multiply the result returned in eax with the saved value of eax# [ P7 q/ C4 j) H6 M4 I- h
pop ebx
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imul eax, ebx* h. p! A( W1 c+ J4 V
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ret
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end_recursion:) }) ~1 }, _5 G- Y# }; }
; Return 1 when eax is 0 (base condition)
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mov eax, 1
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ret

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