微机原理实验，8086汇编

description

编写一个简单的程序：将“This is my first ASM program-姓名（汉语拼音各人的姓名）”放在DS=1000H，BX=0000H开始的存储器单元中，然后将该内容搬移到BX=0100H开始的单元中，最后将该字符串通过DOS功能调用显示在屏幕上。

EXE版本

; 声明数据段
data segment
    msg db "this is first asm program-ya*******ao$"
    db 1000 DUP(0)     ; 未初始化数据
ends

; 声明栈段
stack segment
    dw   128  dup(0)
ends

; 声明代码段
code segment
start:
    mov ax, data        ; 将数据段的首地址mov到ds寄存器，需要注意的是不能直接mov ds, data，必须需要通过ax寄存器间接赋值，这个是汇编的知识。
    mov ds, ax 
    mov ax, 1000h
    mov es, ax    ; 同时将数据段首地址mov到es(extended segment)
    
    mov si, 0
    mov di, 0
    mov cx, 40          ; 40为字符串msg的长度
    
copy1:                     ; 数据从ds复制到1000:0000
    mov ax, [si]
    mov es:[di], ax    ; si和di在没有显式标明段时默认是ds段，这里es:[di]是1000:0000
    add si, 2
    add di, 2
    loop copy1
        
    mov si, 0    
    mov di, 0100h         
    mov cx, 40
    
copy2:                        ; 将数据复制到1000:0100
    mov ax, [si]
    mov es:[di], ax       ; es:[di]为1000:0100
    add si, 2
    add di, 2
    loop copy2

	mov ax, 1000h
    mov ds, ax            ; 按照题目ds为1000
      
    lea dx, msg             ; lea(load effective address，加载有效地址), 将msg的地址存到dx寄存器中，等效于mov dx, offset msg
    mov ah, 9
    int 21h        ; int 21中断，output string at ds:dx
    
    ; wait for any key....    
    mov ah, 1
    int 21h
    
    mov ax, 4c00h ; exit to operating system.
    int 21h    
ends

end start ; set entry point and stop the assembler.

com版本

org 100h

; add your code here 

mov ax, 1000h
mov dx, ax
mov bx, 0000h

mov [02h], 'T'
mov [04h], 'h'
mov [08h], 'i'
mov [0ah], 's'
mov [0ch], ' '
mov [0eh], 'i'
mov [10h], 's'
mov [12h], ' '
mov [14h], 'm'
mov [16h], 'y'
mov [18h], ' '
mov [1ah], 'f'
mov [1ch], 'i'
mov [1eh], 'r'
mov [20h], 's'
mov [22h], 't'
mov [24h], ' '
mov [26h], 'A'
mov [28h], 'S'
mov [2ah], 'M'
mov [2ch], ' '
mov [2eh], 'p'
mov [30h], 'r'
mov [32h], 'o'
mov [34h], 'g'
mov [36h], 'r'
mov [38h], 'a'
mov [3ah], 'm'
mov [3ch], '-'
mov [3eh], 'y'
mov [40h], 'a'
mov [42h], 'n'
mov [44h], 'g'
mov [46h], 's'
mov [48h], 'h'
mov [4ah], 'i'
mov [4ch], 'z'
mov [4eh], 'h'
mov [50h], 'a'
mov [52h], 'o'

mov cx, 40 
mov bx, 0002h

start:
    mov ax, [bx]
    mov [bx+0100h], ax
    add bx, 2
loop start

mov bx, 0102h
mov cx, 40

print:
    mov dl, [bx]
    mov ah, 02h
    int 21h
    add bx, 2
loop print

mov ah, 4ch
int 21h

ret