汇编子程序递归问题

        CODE  SEGMENT

              ASSUME   神纯指 CS:CODE

      START:  MOV       CX,6

              MOV       BX,5

              CALL      DIGUIMI

              CALL      DIGUIAX

              MOV       AH,4CH

              INT       21H

================================

     DIGUIMI  PROC      NEAR

               cx为幂次，bx为底

               ax 中返回 乘幂              

              CMP       CX,0

              JNE       @DIGUIMI

              MOV       AX,1

              RET

   @DIGUIMI:

              DEC       CX

              CALL      DIGUIMI

              MUL       BX

              RET

     DIGUIMI  ENDP

=================================

     DIGUIAX  PROC      NEAR

               递归显示 AX 中的无符号数（10进制）     

              PUSH      BX

              PUSH      DX

              MOV       DX,0

              MOV       BX,10

              DIV       BX

       游配       CMP       AX,0

              JE        @DIGUIAX1

              CALL      DIGUIAX

  @DIGUIAX1:

              OR        DX,30H

              MOV       AH,2

              INT       21H

              POP       DX

           裤好   POP       BX

              RET

     DIGUIAX  ENDP

=================================

        CODE  ENDS

              END       START

#include <stdio.h>int fac(int n)

{

if (n) return fac(n-1)*n

return 1

}void main()

{

printf("%d\n", fac(6)) // 数字随枣消液你改桥好，也可以通凳物过键盘输入。

}

可计算1-10000的阶改备乘野余

DATAS SEGMENT

string db"Please input a number(0-10000):" ,0dh,0ah

wrongstring db" A number please! ",0ah,0dh,'$'

inputbuffer db6,?,6 dup(?)

c10 dw10 输入时进制转换的数

n dw? 要求阶乘的数

m dw? 步长

c dw? 进位颂歼滚

i dw?

outputbuffer dw30000 dup(?)

DATAS ENDS

*******************************************************************************

STACKS SEGMENT

此处输入堆栈段代码

STACKS ENDS

*******************************************************************************

CODES SEGMENT

ASSUMECS:CODES,DS:DATAS,SS:STACKS

START:

MOV AX,DATAS

MOV DS,AX

-------------------------------------------------------

main proc

call input

call fractor

mov cx,di

routput:循环输出

push cx

mov di,cx

call output

pop cx

dec cx

cmp cx,0

jge routput

displaylength:

mov dl,0dh

mov ah,2

int 21h

mov dl,0ah

mov ah,2

int 21h

jmp Start

exit:

mov ah,07h

int 21h

mov ax,4c00h

INT 21H

main endp

-------------------------------------------------------

--------------------------------------------------------

input proc near

lea bx,string

mov cx, 33

disstring:

mov dl,[bx]

mov ah,2

int 21h

inc bx

loop disstring

jmp inputinit

wronginput:

lea dx,wrongstring

mov ah,9

int 21h

inputinit:

lea dx,inputbuffer 装载输入缓冲区首地址

mov ah,0ah 输入功能代码

int 21h 从键盘输入一个数，以回车键结束

mov ax,0累加器清0

mov cl,inputbuffer+1循环次数

mov ch,0

lea bx,inputbuffer+2装载字符存放区首地址

inputone:

mul c10

mov dl,[bx]

cmp dl,'0'

jbwronginput

cmp dl,'9'

jawronginput

and dl,0fh

add al,dl

adc ah,0

inc bx

loop inputone

mov n,ax

mov dl,0dh

mov ah,2

int 21h

mov dl,0ah

mov ah,2

int 21h

ret

input endp

-----------------------------------------------------

fractor proc near

mov cx,n

mov i,1d

mov m,0d

push dx

mov di,0d

mov ax,di

mov bx,2d

mul bx

mov si,ax

pop dx

mov outputbuffer[si],1d

ctrli:

mov c,0

mov di,0d

ctrldi:

cmp di,m

jbe done

jmp cmpc

done:

push dx

mov ax,di

mov bx,2d

mul bx

mov si,ax

pop dx

mov ax,outputbuffer[si]

mov bx,i

mul bx

add ax,c

adc dx,0

mov bx,10000

div bx

mov c,ax

push dx

mov ax,di

mov bx,2d

mul bx

mov si,ax

pop dx

mov outputbuffer[si],dx

inc di

jmp ctrldi

cmpc:

cmp c,0

jathree1

jmp next

three1:

inc m

mov ax,c

mov outputbuffer[si+2],ax

next:

inc i

cmp cx,0

jng if0

loop ctrli

if0:

mov di,m

ret

fractor endp

-----------------------------------------------------

output proc near

C2:

push dx

mov ax,di

mov bx,2d

mul bx

mov si,ax

pop dx

mov bx,outputbuffer[si]

bid proc

mov cx,10000

mov ax,bx

mov dx,0

div cx

mov bx,dx

mov cx,1000

call ddiv

mov cx,100

call ddiv

mov cx,10

call ddiv

mov cx,1

call ddiv

ret

bid endp

---------------------------------

ddiv proc

mov ax,bx

mov dx,0

div cx

mov bx,dx

mov dl,al

add dl,30h

mov ah,02h

int 21h

ret

ddiv endp

ret

output endp

--------------------------------------------

CODES ENDS

END START

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