用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 102 f/ C7 O9 C& K1 [$ A
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section .data ;9 D, y% V- q) o8 J9 l6 E
userMsg db 'Please enter a single-digit number: ' ;
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lenUserMsg equ $-userMsg ;6 ] R: V) z- u3 ` N
disp1Msg db 'The factorial of '
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lenDisp1Msg equ $-disp1Msg / T8 l, K# p( L) M: a8 Z
disp2Msg db ' is '$ I2 S/ R2 [# ?& S9 D K
lenDisp2Msg equ $-disp2Msg 0 S3 k# D0 D" _. t% x+ w! q, Z6 H
errMsg db 'Wrong input. Only one-digit number can be accepted.',0x0a d8 p+ {1 [# ^0 ]& v6 d2 c
lenErrMsg equ $-errMsg & Z, R7 r3 _: C! j0 e' z M
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section .bss ;
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num resb 1
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result resd 1* y* I& k M+ Y% K) }; s
buffer resb LEN_OF_OUTPUT+1. n7 j" R8 @/ W' k) _- [0 E& ~
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section .text ;* r j r3 ~; a c- W% I, R3 s
global _start
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_start: ;$ X7 d \0 J5 U# Z1 D, x, R
mov eax, 4
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mov ebx, 1 F( R {6 m V
mov ecx, userMsg4 j% y1 }8 I3 Q5 q% O- i- {) ^
mov edx, lenUserMsg
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int 80h) W! G: R! C# L$ g$ G" J9 ]
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; read user's input
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mov eax, 3. s7 M" W7 K8 I* W
mov ebx, 0" _, {& g3 S/ W$ f, E8 U
mov ecx, num
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mov edx, 1 ; length of input buffer" g$ L% ]" n8 L9 m+ _) e7 `' d1 @
int 80h
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mov eax, [num]
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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; j2 t$ h3 h- A& X* U5 k
ERROR:& }& N5 k, E6 V7 g$ n
mov eax, 4* I$ N* G* n/ \' K
mov ebx, 1
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mov ecx, errMsg9 \! V- V3 b9 t2 p$ C
mov edx, lenErrMsg
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int 80h $ Z3 D1 L% U% Y
jmp EXIT
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DO_CALC:
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call factorial
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; The parameter is in eax. After calling, the result will be stored in eax.; O% }1 l8 s% D1 c1 r+ n1 ^
mov [result], eax ; Save the result* J. c6 Q6 m6 q8 x) r4 N
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; print result4 ?8 @8 i; U% e& f; |# ]
mov eax, 4" Z8 W7 s! f$ ^- i7 n
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, 41 t1 b0 ]7 J: ~' V
mov ebx, 1% b; A8 ]7 h. K1 k, ^& Y$ ^2 a. G
mov ecx, num
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mov edx, 1& a1 I/ o7 q3 K4 Z3 k; j
int 80h _) q+ Q' K6 J1 q. N
mov eax, 46 N4 p3 I& \* T
mov ebx, 1
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mov ecx, disp2Msg
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mov edx, lenDisp2Msg8 E7 n! `7 w+ E% G
int 80h
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; output number7 {* n5 a/ G- |$ c3 Q
mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY
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mov byte [buffer+LEN_OF_OUTPUT],0x0a3 i x2 ~ C" [! ^3 O
lea esi,[buffer+LEN_OF_OUTPUT]
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mov ebx,10
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ASC_LOOP:
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mov edx,0 ; clear dx prior to dividing edx:eax by bx- ~! I& E% \3 }
div ebx ;DIV EAX/10
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add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER & M$ |# X9 P! ^" `( a! x) y, z. ~
dec esi ; store characters in reverse order- x' X9 N: u8 z B$ _: [8 e
mov [esi],dl" S! S6 z0 ^& x: j" X% V6 J P
cmp eax,0 ( k4 m; ~6 x% U( g
jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII) B0 |% A1 a9 P- h
jmp ASC_LOOP ;ELSE REPEAT5 s( e O- G" C; n! h1 A/ c
END_ASC:
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mov eax, 4
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mov ebx, 11 w8 _6 I- c$ y( q6 U
mov ecx, esi
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lea edx,[buffer+LEN_OF_OUTPUT+1]! ~( Y( X* O+ V; J* V" t$ d+ E
sub edx, ecx
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int 80h
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; Exit code
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EXIT: A3 X# ^; [0 _) z% ?
mov eax, 1
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mov ebx, 04 ^' S* ]2 _$ J1 r/ C- ?3 i2 ^: y# Q
int 80h: d, Z* p7 J* ] r F9 [
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; Factorial function using recursion; g& i# d. S8 G$ U, E
factorial:
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; Check if eax is 0 (base condition)" C3 r# {2 }6 \1 J( o4 X
cmp eax, 08 a" h: F; M4 r5 p7 z" q4 Y: r
jz end_recursion% X9 T) p- E6 A
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; Save the current value of eax
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push eax+ E, p* |' B+ L) }
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; Decrement eax and call factorial function recursively
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dec eax5 \1 E, Y" T: L8 V: ^$ h! C
call factorial
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; Multiply the result returned in eax with the saved value of eax
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pop ebx' Z3 U7 Q" Q! n( a8 V+ X# @
imul eax, ebx" d: x. K1 s7 C9 o6 b, V
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ret+ [4 g& l. A# P0 u( y. d# Q- N
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end_recursion:
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; Return 1 when eax is 0 (base condition)
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mov eax, 15 W0 p/ I/ U) ]5 R- w1 U! ]5 E
ret

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