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: ' ;
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lenUserMsg equ $-userMsg ;
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disp1Msg db 'The factorial of '' P" `7 c# A- \' ^7 {" n! g# j4 M
lenDisp1Msg equ $-disp1Msg 8 [( I! Q$ K6 T* S; T5 `" i
disp2Msg db ' is '
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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 $-errMsg ; f! U; u0 C& P" j' z1 C
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section .bss ;. K' U( \8 d7 \
num resb 13 G, ?- e+ ^! e+ v" l* M( K
result resd 1
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buffer resb LEN_OF_OUTPUT+18 d6 ]0 O! I! R
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section .text ;
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global _start4 `, L: w5 c# i' X' l' o+ n
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_start: ;& T# v7 k1 A9 D z
mov eax, 4; L% |- q3 U# E9 N+ h' p
mov ebx, 16 Y' H$ c f, a7 I, @$ F8 J" P" Z, J
mov ecx, userMsg
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mov edx, lenUserMsg
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int 80h( }: f& T3 A1 R! {6 \* a
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; read user's input
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mov eax, 35 H( Z) Q) O m! d7 @/ @' C! S+ _
mov ebx, 0" F1 E2 C# C' M( w: r3 W, n
mov ecx, num
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mov edx, 1 ; length of input buffer! c Y: h, |6 Z6 `* o! T: c8 \
int 80h( v, ?5 \# ~& N( v. w& r( g4 T
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mov eax, [num]8 H7 D9 h' x% v3 E3 O; t( Y0 u
and eax, 0xFF9 ^4 k' Z+ n+ U5 i- H
sub eax, 0x30
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jl ERROR
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cmp eax, 101 T3 b/ ?5 H" m
jl DO_CALC
ERROR:$ }& `8 S7 F' i' |/ X
mov eax, 4; C8 K! ^8 P7 j* G1 P) B. L4 ]% l
mov ebx, 1
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mov ecx, errMsg6 D4 w4 R& G1 O. \
mov edx, lenErrMsg
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int 80h / D* O3 Y7 M8 m$ P5 ^, W" e5 R! f1 @
jmp EXIT+ A( @; l4 ?* r6 o$ ^1 y- s5 u0 f
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DO_CALC:$ v$ I1 r/ c, K5 V9 b
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 result3 r: K7 M' m! H7 r/ g, G% A& z/ b! X7 a
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; print result% ?3 A0 J, g+ t- ]" Y
mov eax, 45 x" J, ?) w. ]' k7 L- L. e6 k; @
mov ebx, 1
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mov ecx, disp1Msg
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mov edx, lenDisp1Msg
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int 80h
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mov eax, 4& A, I/ g7 H8 `6 B0 w* N8 s* b; B
mov ebx, 1
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mov ecx, num
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mov edx, 1* j) R8 d; O6 G& ?9 Z$ R
int 80h
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mov eax, 4
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mov ebx, 16 X. o7 O* a1 L% u$ E% Y
mov ecx, disp2Msg1 ?. m8 K X! {$ f9 c9 S0 i- E
mov edx, lenDisp2Msg
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int 80h - R9 v* B) F" r9 l* G7 y& }
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; output number8 u, g5 t, v: W u% j H
mov eax, [result] ; VALUE THAT I WANT TO CONVERT AND DISPLAY
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mov byte [buffer+LEN_OF_OUTPUT],0x0a* i; B3 f4 l* q9 |
lea esi,[buffer+LEN_OF_OUTPUT]1 s4 D# p" C% t2 l' k( \
mov ebx,10
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ASC_LOOP:
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mov edx,0 ; clear dx prior to dividing edx:eax by bx7 i* A& Y3 v( B( P/ O
div ebx ;DIV EAX/10' C% |# O2 G/ x( E
add edx,0x30 ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER 2 q8 `/ L7 _2 J6 w8 ?
dec esi ; store characters in reverse order7 X0 t# |- R: q' `) g
mov [esi],dl
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cmp eax,0 0 [. T& ?6 N" J" d3 Z' n, E
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, 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]. M6 ^" `' c. q) M) R# g2 \5 }
sub edx, ecx
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int 80h ! d5 Z: g3 t# U. k+ N
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; Exit code! o* \6 G. L8 f2 G9 a. S ~
EXIT:; j Q4 T8 R, m$ U
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:
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; Check if eax is 0 (base condition)
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cmp eax, 0
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jz end_recursion8 O% _4 y( x; W, J4 \: ^/ e% E
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; Save the current value of eax# z9 W- p) J5 t {, s7 B! ?
push eax5 E6 u3 p; H) D
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; Decrement eax and call factorial function recursively) V. `. u; F- t. S
dec eax
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call factorial
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; Multiply the result returned in eax with the saved value of eax Z a' W3 o. \& b, z
pop ebx
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imul eax, ebx& \+ x* L# q1 G: ^ @* F4 {
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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, 1/ {$ c) n, w$ P$ {
ret

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