用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 10# ^7 K" v2 V; D" w7 K2 ]! p% A
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section .data ;
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userMsg db 'Please enter a single-digit number: ' ;% |- o4 P" g S: ^
lenUserMsg equ $-userMsg ;: L- h, G0 R( _2 _& w
disp1Msg db 'The factorial of '
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lenDisp1Msg equ $-disp1Msg 8 G5 R5 K% ^3 r* ~+ k2 n9 w5 l% T
disp2Msg db ' is '
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lenDisp2Msg equ $-disp2Msg & L9 F, e8 M2 Q! a- |; c. B2 E
errMsg db 'Wrong input. Only one-digit number can be accepted.',0x0a( g" M1 C0 W* ]4 F9 u& P. z" Q
lenErrMsg equ $-errMsg
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section .bss ;8 Q+ c, l0 {8 X1 t4 L+ L9 F# |
num resb 1* s. n" l9 y' g) R) ~! @
result resd 1
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buffer resb LEN_OF_OUTPUT+16 j- L3 C& p0 d* ~
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section .text ;# F+ M" n# H0 W$ \( i$ F8 V
global _start0 S( Y" N W; e1 }" d& s
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_start: ;( c7 V! g6 f, f) b5 @1 C$ z
mov eax, 4" c7 y. Y4 ?; t
mov ebx, 1
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mov ecx, userMsg4 g8 j9 M2 m3 Z0 S* l: `
mov edx, lenUserMsg+ z1 ~) P2 h: x' V& D
int 80h! ]1 z, J c+ t( h& I" p
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; read user's input
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mov eax, 3
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mov ebx, 06 m0 ?0 b) N+ V2 {3 w$ |! P% d
mov ecx, num
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mov edx, 1 ; length of input buffer
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int 80h2 X# h% Z6 z: P X2 D: o3 u! o
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mov eax, [num]
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and eax, 0xFF
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sub eax, 0x30" O, w1 v) ~; Z$ s2 c& _. k1 s
jl ERROR6 ~- K& d$ P6 g q. u! r
cmp eax, 10( z1 j; k' M; s" i
jl DO_CALC( t7 ^1 q8 H& }0 G1 W6 J" e. f
ERROR:
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mov eax, 4. M, h) W1 d" i5 M9 W; z5 n8 i
mov ebx, 1$ o- `! c3 B `# W
mov ecx, errMsg
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mov edx, lenErrMsg' B; h, C3 C3 k% P
int 80h
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jmp EXIT
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DO_CALC:7 U. y+ n* D9 `* B5 M+ `' g7 j
call factorial6 w: G1 [/ F$ U: a6 B
; The parameter is in eax. After calling, the result will be stored in eax.' ~, O* e {8 v) {( n* k
mov [result], eax ; Save the result
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; print result
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mov eax, 49 [, v2 J5 e, ]9 y
mov ebx, 1
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mov ecx, disp1Msg$ Z7 j1 u N: G% L. g6 c4 {
mov edx, lenDisp1Msg6 K: L. ?: u9 ]. J- u0 Q
int 80h 4 `9 ]7 _8 s& x' x5 C5 g# T5 K
mov eax, 4/ t2 Z1 q! v( U% D. i3 E* s
mov ebx, 1
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mov ecx, num7 U- \" D4 D' J# ?$ \1 O
mov edx, 1/ S3 w' }: B5 u2 K& ]
int 80h
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mov eax, 4
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mov ebx, 19 ~8 {: z8 Y2 ` N# A
mov ecx, disp2Msg
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mov edx, lenDisp2Msg% ?6 X" u. |6 ]! j
int 80h
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; output number
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mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY( l) q, A( v% z& P+ o7 S4 f
mov byte [buffer+LEN_OF_OUTPUT],0x0a& N6 q: l0 P" Z( i! x3 t* e4 C$ H
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
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div ebx ;DIV EAX/108 X6 @$ R# z0 ~4 O; _" V/ e
add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER % |" D: c" M5 Q( n
dec esi ; store characters in reverse order
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mov [esi],dl
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cmp eax,0
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jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII
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jmp ASC_LOOP ;ELSE REPEAT/ [" G/ }- Q) E+ B3 i& ]
END_ASC:
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mov eax, 4
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mov ebx, 1
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mov ecx, esi( n, J- _; y* o* j) b1 Y6 H
lea edx,[buffer+LEN_OF_OUTPUT+1]! u3 Q4 n7 ~2 U7 h- G/ n+ k
sub edx, ecx
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int 80h
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; Exit code9 N9 m, H1 E; @, ~6 Z2 f/ s7 d
EXIT:+ x- F3 ^) ?: o
mov eax, 1
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mov ebx, 0
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int 80h$ K z& y3 U4 z3 y
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; Factorial function using recursion
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factorial:
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; Check if eax is 0 (base condition); h* y! B& i) T. J& {* R6 ^. c
cmp eax, 0
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jz end_recursion" }+ X2 w" t9 {$ L7 C) w* K- O
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; Save the current value of eax
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push eax/ g+ O9 ^: C2 q; C" r
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; Decrement eax and call factorial function recursively7 n; X/ m, z3 i c5 C5 t
dec eax
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call factorial2 }, Q h- `! u! Y
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; Multiply the result returned in eax with the saved value of eax. L* _/ s+ W6 L
pop ebx3 t2 i0 I- j1 A: n7 a
imul eax, ebx4 I3 k! v; g9 a7 C7 X
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ret
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end_recursion:
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; Return 1 when eax is 0 (base condition)7 D$ t" \6 W# t8 {# t3 Y
mov eax, 1
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ret

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