用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 10# ~/ }9 x' Q. x
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section .data ;& b* @/ A# d1 \
userMsg db 'Please enter a single-digit number: ' ;6 W3 _* Z* m5 z% I
lenUserMsg equ $-userMsg ;
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disp1Msg db 'The factorial of '
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lenDisp1Msg equ $-disp1Msg 7 e! O, j3 e, E
disp2Msg db ' is '- o4 N) T1 ~# w! ?
lenDisp2Msg equ $-disp2Msg ' ^% k! E. I0 z. d7 g
errMsg db 'Wrong input. Only one-digit number can be accepted.',0x0a: v* W+ m: J' u) U F2 f. E
lenErrMsg equ $-errMsg ) b2 j# F. p- U0 e) U8 d5 g
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section .bss ; |: C z* q; Z9 A. t8 V
num resb 1/ [/ G! W1 i! }0 U- _& M. K/ ]( b
result resd 1
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buffer resb LEN_OF_OUTPUT+16 [" h: S8 q" s' u9 |
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section .text ;& _7 r, x# f$ M: r+ E) Y0 L
global _start& j$ m( ]3 t, x5 g+ L1 F
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_start: ;
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mov eax, 42 B9 r0 _) D V
mov ebx, 1
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mov ecx, userMsg
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mov edx, lenUserMsg0 E- V& _$ ~8 {* M1 @
int 80h
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; read user's input
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mov eax, 3
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mov ebx, 02 s. V5 v& W8 g/ R/ U" ~/ B$ [
mov ecx, num ) @, Q6 M5 o2 c, f; k+ Y0 z: s
mov edx, 1 ; length of input buffer
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int 80h( H/ Q( {1 N2 O4 l& t9 w
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mov eax, [num]
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and eax, 0xFF2 c( X/ n) i3 v! J3 H4 \
sub eax, 0x30
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jl ERROR) Q* o9 N; @9 p0 W( L2 S& Z
cmp eax, 10" A7 l$ x! K" o: }; ]5 K$ z; G
jl DO_CALC
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ERROR:
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mov eax, 4
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mov ebx, 10 z3 t; L6 l; y' A
mov ecx, errMsg
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mov edx, lenErrMsg
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int 80h
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jmp EXIT
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DO_CALC:
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call factorial9 C" [5 A4 N7 U
; The parameter is in eax. After calling, the result will be stored in eax.
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mov [result], eax ; Save the result* }2 I- ]" F+ W4 ]
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; print result
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mov eax, 4% m7 H8 P1 O3 M( f6 Q7 j# F% y. K6 j
mov ebx, 1
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mov ecx, disp1Msg
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mov edx, lenDisp1Msg5 P$ Q3 }8 d9 _; b! u: A8 R
int 80h
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mov eax, 4- F, ` Z' t! o M
mov ebx, 1
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mov ecx, num$ f8 X. F w& S
mov edx, 17 a+ o: t. W: ^
int 80h
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mov eax, 4
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mov ebx, 16 f7 _% s! ~4 Y3 P1 b8 `- y
mov ecx, disp2Msg
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mov edx, lenDisp2Msg m! u0 Z$ A+ c: w; B
int 80h
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; output number
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mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY6 T! E0 |# t( g
mov byte [buffer+LEN_OF_OUTPUT],0x0a9 y2 D U; g& q2 |
lea esi,[buffer+LEN_OF_OUTPUT]
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mov ebx,10 3 J9 r+ I0 o; Y+ H6 _; g
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( \5 J+ h4 D0 O: A
add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER * s( O4 Q) g( I2 h9 B. ^
dec esi ; store characters in reverse order
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mov [esi],dl$ a2 Q+ {& `0 Q I
cmp eax,0
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jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII0 n$ G8 f* R: z3 P( [4 O7 q
jmp ASC_LOOP ;ELSE REPEAT- n! R9 \; e: t E A! F3 g
END_ASC:# T. @. H9 D6 V& H2 M' E! L, n% ^
mov eax, 4% o" ^% }, v9 K# ~5 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 : P# u6 G$ C% T
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; Exit code
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EXIT:! H: ?% p0 K# m& \5 w
mov eax, 17 s6 T; v5 @ c" A5 U+ h, g) E
mov ebx, 0* R/ t$ d9 y' t+ ]5 o, N3 ~
int 80h& l. _$ Q# x6 _9 [! P
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; Factorial function using recursion
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factorial:* H. k5 n+ Z+ W0 \" n
; Check if eax is 0 (base condition)0 Q/ S& h6 B6 X% ~+ [' [0 G; a* Z- h
cmp eax, 0+ `% e$ W7 R( s% Q
jz end_recursion. n6 Y, z. Q/ S. N
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; Save the current value of eax
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push eax$ M- J8 }% b: M7 q5 A$ c. q1 V; G
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; Decrement eax and call factorial function recursively9 R: N. f/ W& @7 F
dec eax' [& Q: ]- n8 A9 D
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- q3 _3 B$ X" |0 ^6 ?! Z& z
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ret& V1 T4 w0 `8 I- v% U: m& ?- U
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end_recursion:; u0 `* {0 z! o/ f
; Return 1 when eax is 0 (base condition)
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mov eax, 1$ Y: {0 l3 K" K( W* a* R" n. Q/ B
ret

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