求n的阶乘,要求用汇编语言,迭代法和递归法

发布网友发布时间：2023-07-21 03:57

共2个回答

热心网友时间：2024-11-30 15:31

递归阶乘：
       CODES  SEGMENT
              ASSUME    CS:CODES
      START:
              MOV       BX,8
              CALL      FACT
              CALL      DISPAX

              MOV       AX,4C00H
              INT       21H

        FACT  PROC      NEAR
              AND       BX,BX
              JZ        FACT1
              PUSH      BX
              DEC       BX
              CALL      FACT        ;递归调用
              POP       BX
              MUL       BX          ;结果在ax中
              RET
      FACT1:  MOV       AX,1
              RET
        FACT  ENDP
;==============================================
      DISPAX  PROC      NEAR
              PUSH      BX
              PUSH      CX
              PUSH      DX
              PUSH      SI
              PUSH      DS
              PUSH      AX
              MOV       AH,2
              MOV       DL,' '
              INT       21H
              POP       AX
              PUSH      CS
              POP       DS
              MOV       BYTE PTR @NZ,0
              PUSH      AX
              LEA       SI,@DIVARR
              MOV       CX,5
        @@1:
              POP       AX
              MOV       DX,0
              MOV       BX,[SI]
              DIV       BX
              PUSH      DX
              CMP       AL,0
              JNE       @@2
              CMP       BYTE PTR @NZ,1
              JE        @@2
              CMP       CX,1
              JE        @@2
              MOV       DL,20H
              JMP       @@3
        @@2:
              ADD       AL,30H
              MOV       DL,AL
              MOV       BYTE PTR @NZ,1
        @@3:
              MOV       AH,2
              INT       21H
              INC       SI
              INC       SI
              LOOP      @@1
              POP       DX
              POP       DS
              POP       SI
              POP       DX
              POP       CX
              POP       BX
              RET
     @DIVARR  DW        10000,1000,100,10,1
         @NZ  DB        0
      DISPAX  ENDP
;===============================================
       CODES  ENDS
              END       START

递推阶乘自己求着试试。

热心网友时间：2024-11-30 15:31

STACK SEGMENT PARA STACK 'STACK'

DW 64 DUP (?)

STACK ENDS

DATA SEGMENT

N DB 8

FUNCN DW ?

DATA ENDS

CODE SEGMENT

ASSUME CS:CODE,SS:STACK,DS:DATA

MAIN PROC FAR

START: PUSH DS

MOV AX,0

PUSH AX

MOV AX,DATA

MOV DS,AX

PUSH CX

MOV AH,0

MOV AL,N

CALL FACTOR

MOV FUNCN,AX

POP CX

RET

MAIN ENDP

FACTOR PROC NEAR

PUSH AX

SUB AX,1

JNE AGAIN

POP AX

JMP FIN

AGAIN: CALL FACTOR

POP CX

MUL CL

FIN: RET

FACTOR ENDP

CODE ENDS

END START

请放心使用

有问题的话请追问

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