用汇编语言实现递归 - 爱吱声

LEN_OF_OUTPUT equ 10' E& n0 B, B" U' w( }4 }
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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 '/ ]7 y: [" @* ~% M8 d2 t( U
lenDisp1Msg equ $-disp1Msg 8 ~* n9 w, R3 N% w! N J3 K
disp2Msg db ' is '
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lenDisp2Msg equ $-disp2Msg g# c% c" B" C# d) R
errMsg db 'Wrong input. Only one-digit number can be accepted.',0x0a
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lenErrMsg equ $-errMsg 5 Y$ r0 \' e) o; b) Z
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section .bss ;% W8 |7 ^& Q3 g( [
num resb 1
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result resd 1! |) U3 h" i1 B, I- n7 [
buffer resb LEN_OF_OUTPUT+1
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section .text ;
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global _start
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_start: ;! ~ D$ E' i+ Y/ R3 {& t9 [
mov eax, 4' O/ U7 s% t ^/ M
mov ebx, 1
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mov ecx, userMsg- n9 t p7 U G( K9 J' u+ h
mov edx, lenUserMsg/ n: N) m- e& v( j/ D% ^& P
int 80h
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; read user's input
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mov eax, 3' d; B. U& e- e8 P
mov ebx, 0
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mov ecx, num : T7 m! D6 h" j9 T; T# f3 N) ?! a
mov edx, 1 ; length of input buffer- e! Z$ h4 _8 `# l. A$ |
int 80h' ?. C! _' K9 D# [. j% C3 O
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mov eax, [num]
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and eax, 0xFF" x; d7 N, F' B. I+ K9 Z5 b$ {
sub eax, 0x30
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jl ERROR
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cmp eax, 10, Y9 L c8 u. B6 N7 K* D/ C* `( Q
jl DO_CALC
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ERROR:; P& f# O& B- q j! N( H
mov eax, 4
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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 EXIT9 p$ d; O$ ~" `9 X* c3 O( p( H1 R
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DO_CALC:) g9 M( Y9 [4 a6 h d0 b
call factorial
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; The parameter is in eax. After calling, the result will be stored in eax.: M: f2 e5 a- b: p+ M: V
mov [result], eax ; Save the result) ] [' n- |0 l0 P
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; print result9 ^. F, p: n5 I! E" ^: ]! P7 c
mov eax, 49 e* }: W5 ~9 i9 w9 N% V
mov ebx, 19 e& |9 F0 f$ w5 w6 y% ?$ G
mov ecx, disp1Msg9 F2 {1 i" F% {6 Q7 Q3 y/ O2 v
mov edx, lenDisp1Msg
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int 80h S) I3 J' P4 S8 y
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
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mov ebx, 1
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mov ecx, disp2Msg/ ^- @ b6 I* h8 ~5 @# M6 l
mov edx, lenDisp2Msg7 q/ [5 y1 l' [5 M8 {1 O
int 80h 9 B& f( _) J y/ f& B x
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; output number1 d6 h* X9 E, O9 q
mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY) U D. X- J/ U9 p F4 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 ; V" r! T8 T% S$ ~1 f
ASC_LOOP:3 `( r1 s& ~% X2 U. V& [
mov edx,0 ; clear dx prior to dividing edx:eax by bx' p, p* v. t( _8 a
div ebx ;DIV EAX/10
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add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER 6 m# W8 d5 Q3 ?! [+ I- U
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; w- M- P: O# D* A* C
jmp ASC_LOOP ;ELSE REPEAT
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END_ASC:
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mov eax, 4
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mov ebx, 1( p; F! t% |" }7 z5 U) D$ [
mov ecx, esi7 r7 Y5 W, x. ]6 j" p2 b: L! Z
lea edx,[buffer+LEN_OF_OUTPUT+1]5 B3 {6 e9 o% I) N& C
sub edx, ecx
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int 80h # w! o- Q; j. z
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; Exit code
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EXIT:
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mov eax, 1. o' U: ~9 O' |, C# V
mov ebx, 0
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int 80h0 ?, C/ f9 q0 B; d& o5 P
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; Factorial function using recursion1 u7 n* `. h7 o) W+ x
factorial:
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; Check if eax is 0 (base condition)3 k: y6 g0 W) x t
cmp eax, 0
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jz end_recursion3 K0 f, V# | j0 Y
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; Save the current value of eax
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push eax3 G) X3 _+ Q E% g- U- O9 K
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; Decrement eax and call factorial function recursively
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dec eax4 a1 H8 L) D5 l6 F* m
call factorial% V. l) [% j) M4 @: E/ ?8 e
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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$ d/ w ?2 O/ I* f& V! K7 j. z. q
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ret3 i2 w: u$ I3 p) Y( c+ d1 X
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end_recursion:
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; Return 1 when eax is 0 (base condition)- b2 h% c- V' ~4 c5 Q" B& u
mov eax, 1
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ret

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