用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 10, z. B% a$ k, ]9 m' ^. C
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section .data ;& y& L) B& r8 z( m$ n# f6 s
userMsg db 'Please enter a single-digit number: ' ;2 c/ N5 j- q9 @* L/ X: F; k
lenUserMsg equ $-userMsg ;% l( B& O( E, j
disp1Msg db 'The factorial of '5 O: q- H, i4 ~( k
lenDisp1Msg equ $-disp1Msg 4 O" u' N1 I% {# Z8 M5 ]
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' V8 Z6 J) a6 j' s
lenErrMsg equ $-errMsg 4 ]4 b7 w8 R' P( U1 B% u
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section .bss ;
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num resb 1
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result resd 1
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buffer resb LEN_OF_OUTPUT+10 W }! `; u5 r: }
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section .text ;7 L! W7 J1 Z$ G$ y s
global _start+ H& M2 \0 Y0 K+ P# |$ ~% J
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_start: ;/ N- u9 ~3 n5 J3 {# |
mov eax, 4, `; G& s+ {. |; _% z1 U
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% r( J0 V3 e7 M6 x9 J
mov eax, 3
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mov ebx, 0
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mov ecx, num / u( ~) H/ A4 a4 U7 G$ Z% M# r" w; u
mov edx, 1 ; length of input buffer8 c0 T! ?8 ~; N! h( k$ y) |
int 80h5 r3 _* N0 _. r$ c5 U6 U: `4 D! m
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mov eax, [num]/ t& U. O6 e, Z2 ~
and eax, 0xFF
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sub eax, 0x30
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jl ERROR
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cmp eax, 10
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jl DO_CALC+ i' G- c9 w# f9 a4 \
ERROR:* K) Q$ e u, {
mov eax, 4
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mov ebx, 15 w9 s+ P( A- p1 J
mov ecx, errMsg
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mov edx, lenErrMsg
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int 80h / `9 X5 g/ C( H3 r4 | w% V
jmp EXIT! t0 s$ z% m4 [- G3 E
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DO_CALC:3 \- ` U! X$ p5 T* g
call factorial/ `; p# f& E6 R, t# d# L4 k& l
; The parameter is in eax. After calling, the result will be stored in eax.0 y) q5 p3 a' ?" ]
mov [result], eax ; Save the result' Z. @) o* X5 G
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; print result
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mov eax, 46 _& w: O I" N
mov ebx, 1, w6 q% k B2 t
mov ecx, disp1Msg, H- |! E( j4 P5 D- o, A
mov edx, lenDisp1Msg
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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, num, `2 O4 ?, x1 D8 S) s/ x
mov edx, 1# ]5 V+ @3 C `2 @* L
int 80h # e# y; j; I* u
mov eax, 41 s9 g1 \" J1 k- U
mov ebx, 1
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mov ecx, disp2Msg
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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
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mov byte [buffer+LEN_OF_OUTPUT],0x0a
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lea esi,[buffer+LEN_OF_OUTPUT]
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mov ebx,10 ! M& i6 ]4 J2 _: C: y3 [2 R/ p! }( R+ 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% p4 E6 \1 V+ V0 ~) R4 }- {3 Q
add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER 0 q( A- J* _# W* j0 z
dec esi ; store characters in reverse order) q& r! X( ]0 j9 Y
mov [esi],dl# M1 [! o8 m1 J- W4 g. a
cmp eax,0 ; n4 v1 S8 h# R6 M
jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII# ^9 @. x- B8 }+ Y/ S
jmp ASC_LOOP ;ELSE REPEAT& [! e' E1 Q/ B" Y- m9 v) z
END_ASC:! e# I2 w) h0 D" H9 Q
mov eax, 4: \9 `3 s+ k: n0 p7 q' B7 x5 F( j
mov ebx, 1
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mov ecx, esi
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lea edx,[buffer+LEN_OF_OUTPUT+1]
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sub edx, ecx
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int 80h ! \2 U A3 t1 f
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; Exit code" k# K8 H+ L$ C0 Z
EXIT:1 `8 e4 t+ E4 E$ e Y3 ?
mov eax, 13 D; G4 g. p+ O
mov ebx, 0
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int 80h
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; Factorial function using recursion5 H ~5 Z( O* E/ o% n- c
factorial:
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; Check if eax is 0 (base condition)
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cmp eax, 0: T% ~! J8 @( f0 n3 A# K
jz end_recursion
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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 recursively. I8 D8 x5 x A0 g
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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ret4 z# U: ?) H3 A9 [. k
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end_recursion:
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; Return 1 when eax is 0 (base condition)
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mov eax, 1
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ret

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