汇编2

最新推荐文章于 2024-11-27 22:54:15 发布

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#汇编程序

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本文通过多个实例展示了8086汇编语言的应用技巧，包括字符串操作、内存寻址、条件跳转等核心概念，适用于初学者快速上手。

1、SI–ds:[si]
DI–es:[di]

assume cs:code,ds:data

data segment
    db 'welcome to names'
    db '----------------'
data ends

stack segment stack
        dw 0,0,0,0,0,0,0,0
        dw 0,0,0,0,0,0,0,0
stack ends

code segment
    start:
            mov ax,stack
            mov ss,ax
            mov sp,32;  栈大小

            mov ax,data
            mov ds,ax
            mov es,ax

            mov si,0
            mov di,16

            mov cx,8
/------------------------------------
copydat：   mov dl,ds:[si];      
            mov es:[di],dl
            add si,2
            add di,2
            loop copydat
-------------------------------------
copydat：   push ds:[si]
            pop  es:[di]
            add si,2
            add di,2
            loop copydat
-------------------------------------/
            mov ax,4C00H
            int 21H
code ends
end start

2、ds:[bx+si] ds:[bx+di]
ds:[bx+si+2] ds:[bx+di+3]

 案例：
   data段每个单词头字母大写

assume   cs:code,ds:data,ss:stack

data segment
       ;'123456789abcdef'
    db  '1.file         '
    db  '2.edit         '
    db  '3.search       '
    db  '4.view         '
data ends

code segment
        start:
        mov ax,data
        mov ds,ax
        mov es,ax

        mov bx,0
        mov si,3

        mov cx,4

upletter: mov dl,ds:[bx+si]
          and dl,11011111B
          mov es:[bx+si],dl
          add bx,16
          loop upletter


        mov ax,4c00H
        int 21H
code ends
end start

3、将单词改为大写

assume cs:code,ds:data,ss:stack

data segment
       ;'123456789abcdef'
    db  '1.ibm          '
    db  '2.doc          '
    db  '3.jkc          '
    db  '4.vax          '
data ends

stack segment
    dw 0,0,0,0,0,0,0,0
    dw 0,0,0,0,0,0,0,0
stack ends

code segment
            start:
            mov ax,stack
            mov ss,ax
            mov sp,32

            mov ax,data
            mov ds,ax

            mov cx,4;   4条数据
            mov bx,0    
            mov si,3;   偏移量

upletter:   push cx
            push si
            mov  cx,3

upRow:      mov dl,ds:[bx+si]
            add dl,11011111B
            mov es:[bx+si],dl
            inc si
            loop upRow

            pop si
            pop cx
            add bx,16
            loop upletter


            mov ax,4c00H
            int 21H
code ends

end start

4、*:mul /:div
OFFSET:
案例：

data segment
    db 256 dup(0)
data ends

stack segment stack
    db 128 dup(0)
stack ends

code segment
        start:
            mov ax,stack
            mov ss,ax
            mov sp,128
        s:  
            mov ax,bx
            mov si,OFFSET s
            mov di,OFFSET s0

            mov dx,cs:[si]
            mov cs:[di],dx
        s0:
            nop
            nop

        mov ax,4c00H
        int 21H
code ends
start end

5、使用jmp指令执行后，cs,ip指向程序的第一条指令

data segment
    db  0,0,0,0,0
data ends

code segment
       start:
             mov ax,data
             mov ds,ax
             mov bx,0
             jmp word ptr ds:[bx+1]

        mov ax,4c00H
        int 21H
code ends
end start
*****************************
data segment
    dd  12345678H
data ends

code segment
       start:
             mov ax,data
             mov ds,ax
             mov bx,0

             mov word ptr ds:[bx],0
             mov ds:[bx+2],cs

             jmp dword ptr ds:[0]

        mov ax,4c00H
        int 21H
code ends
end start

6、jcxz指令，在内存2000H,找值为0，将偏移地址放入
dx

assume cs:code

code segment
    start:
            mov ax,2000H
            mov ds,ax
            mov bx,0

        s:  
            mov cx,0
            mov cl,ds:[bx]
            jcxz ok
            inc bx
            jmp s
        ok:
            mov dx,bx

            mov ax,4c00H
            int 21H
code ends
start end

7、int
data segment
    db 256 dup(0)
data ends

stack segment stack
    db 128 dup(0)
stack ends

code segment
        start:
            mov ax,stack
            mov ss,ax
            mov sp,128

            call cpy_new_int

            mov ax,4c00H
            int 21H
;------------------------------
new_int:
            mov ax,1000H
            mov ax,1000H
            mov ax,1000H
            mov ax,1000H
new_int_end:    nop

;------------------------------
cpy_new_int:
            mov bx,cs
            mov ds,bx
            mov si,OFFSET new_int

            mov bx,0
            mov es,bx
            mov di,7E00H

            mov ex,OFFSET new_int_end
            ret

code ends
start end