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: ' ;' F, s2 X! G; }% V6 K( x) _( T
lenUserMsg equ $-userMsg ;& r1 m8 K/ {' q- t
disp1Msg db 'The factorial of '
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lenDisp1Msg equ $-disp1Msg ! P! w y, M5 B
disp2Msg db ' is ') u4 ]" n5 t; S% g* J- H
lenDisp2Msg equ $-disp2Msg
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errMsg db 'Wrong input. Only one-digit number can be accepted.',0x0a& N( Q9 \; f0 Z
lenErrMsg equ $-errMsg ; ~9 z! i0 n( R' |7 _' N
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section .bss ;+ i8 ]# ?. Z6 d4 f2 y- F: G
num resb 1+ {# Z) S# _6 m. o* Z# b; P, w$ X
result resd 1
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buffer resb LEN_OF_OUTPUT+15 G& b3 ?2 F! s# _
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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, 17 \+ ?* q2 K' @0 d; ?5 E9 C
mov ecx, userMsg B+ i! B0 n* t# Y& O1 S% h
mov edx, lenUserMsg6 |9 E& @5 r8 K3 U3 _# _0 T
int 80h
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; read user's input c; Q- R3 Z e& c; o
mov eax, 3
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mov ebx, 0 a9 x, }; A6 L
mov ecx, num
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mov edx, 1 ; length of input buffer
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int 80h2 [. q7 s8 Z' c
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mov eax, [num]
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and eax, 0xFF
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sub eax, 0x30! m0 o) v1 d' ]- r8 |* ?" n" M: r$ h
jl ERROR6 ^) {! ]) e8 {' H/ W* \- Z
cmp eax, 10+ X* o% X: m" i. }- E2 ~
jl DO_CALC" y" d6 d- H4 {/ n" }! C
ERROR:& G0 @, Z3 z N/ f, a
mov eax, 4
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mov ebx, 1
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mov ecx, errMsg$ C: d* j0 B. L/ }# k% Z
mov edx, lenErrMsg
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int 80h 8 `) X b% P5 q( x
jmp EXIT% Y* a, o/ Z& Q7 w8 c. m" e
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DO_CALC:
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call factorial
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; 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' @: h8 |: w% Z; _( F/ W2 J) Q6 m
mov eax, 4" O! ?2 [% Q( T# }* f$ g) d. {% G
mov ebx, 1
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mov ecx, disp1Msg
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mov edx, lenDisp1Msg& |5 a( m. V' _5 v3 S6 s! |
int 80h
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mov eax, 4
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mov ebx, 1: p3 N% u5 U0 v
mov ecx, num
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mov edx, 1/ `' e1 u: w8 h- ?
int 80h
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mov eax, 4
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mov ebx, 1, t! }* N- d& v6 Q+ ]
mov ecx, disp2Msg1 U+ ]3 i! i0 e3 y7 D, f) n9 Q# f
mov edx, lenDisp2Msg8 v% N: E6 x1 J( T; x8 N4 Y
int 80h 3 V) i# p! L1 s+ h9 p* \0 k
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; output number
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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]; \3 E; T& Z$ e, D, C
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/10- \. a" N0 u. S" Q
add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER
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dec esi ; store characters in reverse order- w! }; g' y* H" H; e
mov [esi],dl
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cmp eax,0 9 l7 @$ i6 H- d& `, I
jz END_ASC ;IF AX=0, END THE PROCEDURE OF NUMBER TO ASCII
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jmp ASC_LOOP ;ELSE REPEAT/ M G+ a5 q) B
END_ASC:7 ]3 k9 Z5 g+ j" I
mov eax, 4
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mov ebx, 1; U; ^! S8 b3 [& ?3 d- K
mov ecx, esi
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lea edx,[buffer+LEN_OF_OUTPUT+1]
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sub edx, ecx. h1 O; V1 W5 H% I; l
int 80h
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; Exit code
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EXIT:! m1 P" d# E6 K* N" I" `4 j* `( U
mov eax, 1
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mov ebx, 06 B; ?7 t4 o$ N% B3 M7 [% {
int 80h% S: p! B( ^1 ~* ]
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; Factorial function using recursion
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factorial: q* d) `0 }0 J% J- s0 Y. `
; Check if eax is 0 (base condition)
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cmp eax, 08 ?1 q0 j# c+ {
jz end_recursion
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; Save the current value of eax* ^3 S' w/ C# C" i# _
push eax
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; Decrement eax and call factorial function recursively
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dec eax
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call factorial' x" T+ ^+ W2 j7 d* G8 `* t* W
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; Multiply the result returned in eax with the saved value of eax" H" _' d3 U4 _. Z+ W/ y
pop ebx
imul eax, ebx2 y4 [( t# B9 C- T* w3 R
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ret
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end_recursion:" Y1 n2 a# P9 Z" x+ ^2 Z3 y
; Return 1 when eax is 0 (base condition)
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mov eax, 1- s/ [+ ^8 V# e x4 g
ret

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