用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 10' d3 O% B, @9 k
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section .data ;
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userMsg db 'Please enter a single-digit number: ' ;
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lenUserMsg equ $-userMsg ;
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disp1Msg db 'The factorial of '
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lenDisp1Msg equ $-disp1Msg # W" c2 h. g1 ~' m0 q
disp2Msg db ' is '$ s# y' M0 b; L* a1 j, E: @1 z$ T- H
lenDisp2Msg equ $-disp2Msg
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errMsg db 'Wrong input. Only one-digit number can be accepted.',0x0a
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lenErrMsg equ $-errMsg % Y( ^% W1 b- a9 S
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section .bss ;
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num resb 1/ K- k) H; Q3 Y
result resd 1
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buffer resb LEN_OF_OUTPUT+17 l, @* ^3 G1 k9 b/ T$ Y
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section .text ;
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global _start
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_start: ;9 W" I( T4 a/ T/ P( N
mov eax, 4
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mov ebx, 1
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mov ecx, userMsg
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mov edx, lenUserMsg
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int 80h8 Q3 W& H6 P1 d( ~6 ~
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; read user's input
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mov eax, 3$ S) o7 u3 z* U3 Q. O6 K/ u
mov ebx, 0
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mov ecx, num
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mov edx, 1 ; length of input buffer4 K: J9 U' L% n" d1 A! d
int 80h+ ?1 z G" F4 J E3 [
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mov eax, [num]
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and eax, 0xFF
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sub eax, 0x30 i+ l1 e7 q1 X% T6 {+ I4 s6 j
jl ERROR
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cmp eax, 10$ Q; p% n/ T, w2 w8 k: x6 e
jl DO_CALC
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ERROR:
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mov eax, 4
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mov ebx, 1) E) g2 ~/ z% v
mov ecx, errMsg2 W. g! u. F/ j' c- u3 L5 P# ?
mov edx, lenErrMsg0 i, P D K, }3 g* f
int 80h ( j% }* R7 Y+ N' b
jmp EXIT
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DO_CALC:* ?9 R' k1 H. X& V. D, |* y3 ]. P9 g- ]
call factorial
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; The parameter is in eax. After calling, the result will be stored in eax.( {) z y6 D! U5 Q3 K
mov [result], eax ; Save the result6 `3 x: b4 J: J
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; print result
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mov eax, 4
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mov ebx, 1. _, L7 [3 W" |1 p9 p1 h6 V$ K
mov ecx, disp1Msg% H- |! A7 n& C( J2 G) b6 w9 y5 `
mov edx, lenDisp1Msg9 l1 y7 p# k4 A& z7 `* v
int 80h
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mov eax, 4
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mov ebx, 1) ?# C5 j% A( Y9 X+ P6 F* Y, {
mov ecx, num
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mov edx, 1
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int 80h
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mov eax, 4
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mov ebx, 1+ u- T7 f" R4 ^+ ~$ m1 p0 K/ T0 s
mov ecx, disp2Msg8 N. e6 A$ V3 j4 {% ]
mov edx, lenDisp2Msg
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int 80h
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; output number5 N |4 T, K) P9 t- D
mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY
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mov byte [buffer+LEN_OF_OUTPUT],0x0a
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lea esi,[buffer+LEN_OF_OUTPUT]
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mov ebx,10 7 n* F* w2 m6 Z. ^. r0 T1 w0 }
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- M" N, l7 q5 M) t7 @
add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER
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dec esi ; store characters in reverse order* \3 o( n. ?1 T" H: Q8 E! {* V
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) s" [7 c0 \$ D3 e: C+ z
END_ASC:
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mov eax, 4
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mov ebx, 1
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mov ecx, esi
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lea edx,[buffer+LEN_OF_OUTPUT+1]9 l4 E) Q" {7 h/ T. }
sub edx, ecx
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int 80h ( @& }- s0 w, G" O4 ~& M* ~
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; Exit code
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EXIT:2 z% i k8 P# c$ ~" u6 a
mov eax, 1
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mov ebx, 0
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int 80h
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; Factorial function using recursion
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factorial:/ B" x1 n/ y0 r# y3 _2 @2 x
; Check if eax is 0 (base condition)- A6 X1 m/ w$ e, x* [) p
cmp eax, 09 {4 k2 z5 X- `6 Z
jz end_recursion
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; Save the current value of eax2 Q( k# {3 z0 M0 c! `
push eax) [: c+ s+ D; K) }7 T* W
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; Decrement eax and call factorial function recursively
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dec eax9 h! {$ w, x% y- x8 x8 g& h3 ]
call factorial
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; Multiply the result returned in eax with the saved value of eax$ R8 f. F) q/ @" `6 j) T6 `3 ?
pop ebx7 z+ A$ |; G o) J
imul eax, ebx5 d5 F g$ R" _1 c- y
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ret3 S7 A& p; C7 n' o* F8 J/ U8 r
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end_recursion:) s! d7 m2 q( r c! g) E: [3 ]
; Return 1 when eax is 0 (base condition)
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mov eax, 1: P& Y2 k) U# o& i2 M6 v" ~
ret

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