数据加密.asm

汇编语言课程设计-数据加密

key=179
data segment
buf1  db  20
      db  ?
      db  20 dup (0)
inf_1  db 0ah,0dh,'enter the plain text:  $'
inf_2  db 0ah,0dh,'cipher          text:  $'
inf_3  db 0ah,0dh,'decrypted           :  $'
inf_4  db 0dh,0ah,'Do you want to decrypt?(Y/N) $'
data ends

stack segment 
 db 100 dup (?) 
stack ends


code segment 
assume cs:code,ds:data


start:
          mov ax,data
          mov ds,ax



     lea dx,inf_1
     mov ah,09h
     int 21h
      
     lea dx,buf1
     mov ah,0ah
     int 21h

     mov cl,buf1+1
     mov ch,0
     mov si,2
     
K1:  xor buf1[si],key
     inc si
     loop K1
     
     lea dx,inf_2
     mov ah,09h
     int 21h

     mov bl,'$'
     mov [si],bl
     lea dx,buf1
     add dx,2
    mov ah,09h
     int 21h

     lea dx,inf_4
     mov ah,09h
     int 21h

 mov ax,0100H ;调用中断21H的01功能,接受一个字符输入并回显
int 21h
 cmp al,'n' ;判断输入字符是否为y或Y,如果是,则继续执行，否则退出
 jz exit
 cmp al,'N'
 jz exit


mov ah,07h
int 21h



    
     mov cl,buf1+1
     mov ch,0
     mov si,2
K2:  xor buf1[si],key
     inc si
     loop K2

     
     lea dx,inf_3
     mov ah,09h
     int 21h

     mov bl,'$'
     mov [si],bl
     lea dx,buf1
     add dx,2
     mov ah,09h
     int 21h

exit:
mov ah,07h
int 21h

mov ah,4ch
int 21h
code ends
end start