用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 10. c( H, r& u8 w0 z( r$ O
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section .data ;" A, @# w5 k& c) L4 l* P/ T
userMsg db 'Please enter a single-digit number: ' ;( @% S/ c! m3 @! I3 B
lenUserMsg equ $-userMsg ;
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disp1Msg db 'The factorial of '! f7 c% Y4 |5 K% e
lenDisp1Msg equ $-disp1Msg ' {/ J. ? Q9 ~* c$ L# Z
disp2Msg db ' is '
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lenDisp2Msg equ $-disp2Msg * V9 v: z; k- k, o0 ^; ]! @8 E
errMsg db 'Wrong input. Only one-digit number can be accepted.',0x0a
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lenErrMsg equ $-errMsg
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section .bss ;
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num resb 1) X k+ [1 w% c$ @
result resd 1) `* a5 b; f1 n7 b) M0 I! g
buffer resb LEN_OF_OUTPUT+1
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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, 1
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mov ecx, userMsg. B3 v, }! U9 J! K" h& f4 f' E2 p' H
mov edx, lenUserMsg
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int 80h
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; read user's input
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mov eax, 3" j; x( L9 O8 ?: b$ X: V( B0 [ Z+ @
mov ebx, 0( f4 d: T% s! r+ D. m* Z
mov ecx, num
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mov edx, 1 ; length of input buffer
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int 80h
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mov eax, [num]7 ^& d9 Z$ U: l" F; n. P3 S
and eax, 0xFF9 W7 n, i i. O Q0 @, M$ ^0 ?
sub eax, 0x30+ ^# |8 U: {) v' T
jl ERROR
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cmp eax, 10
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jl DO_CALC. ]/ Y7 q, u- S% L% z
ERROR:
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mov eax, 4; g' L9 i# {# S' g$ Y
mov ebx, 1
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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:/ u8 y, i* k" U* [3 U9 H
call factorial4 D- h4 d% i; P8 O" ^# _; H; C& f
; The parameter is in eax. After calling, the result will be stored in eax.
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mov [result], eax ; Save the result
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; print result
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mov eax, 43 H/ ^8 ]. S# E, u! V" I# a/ J
mov ebx, 1
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mov ecx, disp1Msg" G% A' E, u. U/ U! g
mov edx, lenDisp1Msg
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int 80h % j1 A: `; V- u; z7 Z% \: n6 y! ^. t
mov eax, 4
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mov ebx, 1
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mov ecx, num
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mov edx, 1
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int 80h
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mov eax, 4& V0 c" g. W E7 _( O0 V+ Y% c/ L
mov ebx, 1# w. B3 Z1 ]8 g
mov ecx, disp2Msg6 a/ n6 S9 N% ^% d* [8 `
mov edx, lenDisp2Msg
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int 80h # @( B$ E! n0 b9 H6 e2 | F3 p
; z$ W. e9 s: o: l$ W" ~8 l
; output number5 R X5 v- {% F+ m! h) I, d
mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY
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mov byte [buffer+LEN_OF_OUTPUT],0x0a% v0 Y! w0 z; h, |' T# ?
lea esi,[buffer+LEN_OF_OUTPUT]- E8 R$ p$ g( ]: z, k
mov ebx,10 ( f7 h; R' H4 [6 u" _
ASC_LOOP:- \9 K1 h& [+ Z& @4 r0 i1 b
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
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dec esi ; store characters in reverse order
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mov [esi],dl. G( L) n% E7 Y9 l
cmp eax,0 $ L/ f& f# e5 U" O+ l
jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII
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jmp ASC_LOOP ;ELSE REPEAT
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END_ASC:
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mov eax, 41 e( t3 ~$ d, L3 G2 }8 U, _
mov ebx, 1! X0 ^7 E4 ^# d
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
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; Exit code
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EXIT:
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mov eax, 1
mov ebx, 0+ y2 @3 v" V4 D& X1 l8 `# U5 r6 x; n
int 80h' g8 k6 j6 d& {- e% S
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; Factorial function using recursion
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factorial:3 C$ z/ j& T& u' l
; Check if eax is 0 (base condition); ~# Z Z7 W6 e4 \. G' y" Q: s3 i
cmp eax, 0! a& O9 {+ Q2 }2 R
jz end_recursion
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; Save the current value of eax3 z) T; Q- v( R/ ^! a6 ?5 B5 A
push eax1 D0 x j6 g# M, B. h
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; Decrement eax and call factorial function recursively! P* J( A) _9 K8 ^( C0 h( R1 Q
dec eax: Q( x( n# F" X) K9 ~0 n0 o
call factorial
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; I$ c$ w3 u. d
; Multiply the result returned in eax with the saved value of eax
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pop ebx' a& c# ^' L* F. F, ^( w y% y
imul eax, ebx
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ret: G5 F: c: b6 K- P, U* v
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end_recursion:7 e9 t }- t- z( b3 @
; Return 1 when eax is 0 (base condition)+ B9 m, M0 c% ]- ?& ~
mov eax, 1) d$ X2 `6 u% x) T( Y
ret

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