用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 10
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section .data ;
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userMsg db 'Please enter a single-digit number: ' ;4 w9 _2 b9 W- p2 H$ F$ b3 t. {: g
lenUserMsg equ $-userMsg ;- u+ H; d ^+ r' U1 W
disp1Msg db 'The factorial of '& ?4 ?0 i' h1 W5 |, F
lenDisp1Msg equ $-disp1Msg 5 U- y5 C3 K. P3 R
disp2Msg db ' is '; R# b8 u1 P, _. ~( J
lenDisp2Msg equ $-disp2Msg
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errMsg db 'Wrong input. Only one-digit number can be accepted.',0x0a5 z* q: _) R7 X$ C3 i, b; g
lenErrMsg equ $-errMsg
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section .bss ;( D' b& I( U6 i* {6 G) ?. ^5 v
num resb 1
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result resd 1
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buffer resb LEN_OF_OUTPUT+1
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section .text ;
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global _start0 {, C$ Y' \0 `3 K$ P5 R0 R
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_start: ;! A7 z- u8 O+ O, }# `7 r; b: \
mov eax, 4 Z, T# j( S% @+ [
mov ebx, 1
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mov ecx, userMsg c3 m" Q$ h3 a( d$ H N- t, f. H
mov edx, lenUserMsg
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int 80h7 N5 ]) w4 Y I% i, L, @8 ~
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; read user's input" p! e. V: D) M5 ~0 ]
mov eax, 3' n: p9 V! s4 ?& `+ @) N& j
mov ebx, 0
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mov ecx, num " q* [- U w, x5 ?( ~* m- l
mov edx, 1 ; length of input buffer. u( s, U& y- V( t3 v
int 80h* \6 q |" i3 S5 N1 ^$ H' Y) ?
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mov eax, [num]1 w& x" u7 x0 r8 ]. |, A: }. t
and eax, 0xFF
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sub eax, 0x30
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jl ERROR: I: \! |8 U: }. y2 p, ]
cmp eax, 10
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jl DO_CALC$ j( K) d$ l6 E" Z, T
ERROR:
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mov eax, 4( m7 {; _- o5 U) j9 m7 a1 @: |
mov ebx, 1! t+ I3 g0 `1 L, E5 {5 @
mov ecx, errMsg
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mov edx, lenErrMsg+ ^+ |" f5 H8 V
int 80h 8 `; T: [. R3 Z
jmp EXIT
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DO_CALC:: ?. D8 o9 |. ^5 Y$ A8 K
call factorial+ i! j' ~9 ?( m% N3 k
; The parameter is in eax. After calling, the result will be stored in eax.
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mov [result], eax ; Save the result3 }# ^% F9 ^4 B0 t
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; print result S/ ?4 R7 j( } b0 u
mov eax, 4( f& J/ u2 z% v& s( {" X3 r* f
mov ebx, 17 I7 ?( x0 y9 {' `- q
mov ecx, disp1Msg
mov edx, lenDisp1Msg
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int 80h , [7 a) q+ y5 E4 K% L
mov eax, 4. b1 `8 J% M% v- [
mov ebx, 1
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mov ecx, num$ K/ Q* f' U' i" C
mov edx, 10 Q+ A5 H3 A4 p9 u8 k
int 80h # s) v! h' ?& L/ G+ z% w2 O# k
mov eax, 4
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mov ebx, 1
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mov ecx, disp2Msg9 k6 c& W4 h- r* n( D
mov edx, lenDisp2Msg! D7 y% y* o" b3 ^( [/ @$ n
int 80h
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; output number
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mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY3 R4 {* E6 ?& |4 E! ~# o9 U3 d' ?
mov byte [buffer+LEN_OF_OUTPUT],0x0a
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lea esi,[buffer+LEN_OF_OUTPUT]
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mov ebx,10 * g* T# B8 I: c' v% a
ASC_LOOP:5 h/ W" f- p6 Q) v* H- p
mov edx,0 ; clear dx prior to dividing edx:eax by bx/ Q: g& H" p5 V7 Q/ X
div ebx ;DIV EAX/109 O! b! Y3 Z8 x+ T! p0 @
add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER # y, Z" a* w* d# {( T! l& ^. m
dec esi ; store characters in reverse order0 X3 M E0 P( d8 S2 X
mov [esi],dl
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cmp eax,0 + X" S# z3 D! \2 W0 x
jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII
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jmp ASC_LOOP ;ELSE REPEAT0 i2 g$ L4 C3 t/ I7 R
END_ASC:
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mov eax, 4! d' X: e) x& X
mov ebx, 1
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mov ecx, esi
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lea edx,[buffer+LEN_OF_OUTPUT+1]; B. |+ k, |4 ~2 D3 e* k5 X
sub edx, ecx ]; Q) A2 E' i c
int 80h , H* U. }' A- u
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; Exit code' m2 ^8 [7 B* l( p1 g( _
EXIT:
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mov eax, 1
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mov ebx, 0
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int 80h* w2 i9 P! ?& v# i
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; Factorial function using recursion6 r) g: f9 v* q$ o2 ]& B
factorial:, }, i) s4 K" s ?
; Check if eax is 0 (base condition)* l% h# _7 t% O5 M8 [
cmp eax, 0
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jz end_recursion! J! I' @) c* T7 w! d* A2 m
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; Save the current value of eax. R% M! e! r3 I3 f* g
push eax; U7 k; I: i0 O) O) T
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; Decrement eax and call factorial function recursively
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dec eax
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call factorial3 B# c: p2 R) {4 W c3 U
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; Multiply the result returned in eax with the saved value of eax* i( l y! n# }8 l
pop ebx
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imul eax, ebx
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ret
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end_recursion:
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; Return 1 when eax is 0 (base condition)
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mov eax, 12 P7 G, L; [/ V
ret

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