实验3 转移指令跳转原理及其简单应用编程

任务一

assume cs:code, ds:data

data segment
    x db 1, 9, 3
    len1 equ $ - x

    y dw 1, 9, 3
    len2 equ $ - y
data ends

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

    mov si, offset x
    mov cx, len1
    mov ah, 2
 s1:mov dl, [si]
    or dl, 30h
    int 21h

    mov dl, ' '
    int 21h

    inc si
    loop s1

    mov ah, 2
    mov dl, 0ah
    int 21h

    mov si, offset y
    mov cx, len2/2
    mov ah, 2
 s2:mov dx, [si]
    or dl, 30h
    int 21h

    mov dl, ' '
    int 21h

    add si, 2
    loop s2

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

 

 

 

（1）

程序运行到loop s1时，指令寄存器IP=001B指向下一条指令，而001B+00F2=010D，高位舍弃，即IP指向000D

（2）

位移量是：16，Loop指令结束地址为：0039，s1指令开始地址为0029，001B-000D=16

or dl, 30h命令占三个字节，其他指令各占两个字节，一共16字节。

 

任务二

assume cs:code, ds:data

data segment
    dw 200h, 0h, 230h, 0h
data ends

stack segment
    db 16 dup(0)
stack ends

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

    mov word ptr ds:[0], offset s1
    mov word ptr ds:[2], offset s2
    mov ds:[4], cs

    mov ax, stack
    mov ss, ax
    mov sp, 16

    call word ptr ds:[0]
s1: pop ax

    call dword ptr ds:[2]
s2: pop bx
    pop cx

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

（1）

ax为s1的偏移地址0021，bx为s2的偏移地址0026，cx为cs即076C。

（2）

由图可以看出AX=s1的偏移量 BX=s2的偏移量，CX则是code段的段地址

 

任务三

assume cs:code ds:data
data segment
x db 99, 72, 85, 63, 89, 97, 55
len equ $- x
data ends
code segment
start:
mov bx,0
mov ax,data
mov ds,ax
mov cx,len
s:
call printNumber
call printSpace
loop s

mov ah,4ch
int 21h

printNumber:
mov al,[bx]
inc bx
mov ah,0
mov dl,10
div dl
OR al,30h
OR ah,30h
mov dh,ah
mov ah,2
mov dl,al
int 21h

mov ah,2
mov dl,dh
int 21h
ret

printSpace:
mov ah,2
mov dl,' '
int 21h
ret

code ends
end start

 

 任务四：

assume cs:code, ds:data

data segment
str db 'try'
len equ $ - str
data ends

code segment
start:
mov cx,len
mov ax,data
mov ds,ax
mov si,0
mov bl,2
mov bh,0
call printStr

mov cx,len
mov si,0
mov bl,4
mov bh,24
call printStr

mov ah,4ch
int 21h

printStr:
mov dl,bh
mov dh,0
mov ax,160
mul dx
mov di,ax
mov ax,0b800h
mov es,ax

mov ah,bl
s:
mov al,ds:[si]
inc si
mov es:[di],ax
add di,2
loop s
ret

code ends
end start

 

 

任务五

assume cs:code,ds:data
data segment
stu_no db '201983290207'
len=$ - stu_no
data ends
code segment
start:
mov ax,data
mov ds,ax
mov di,0
mov ax,0b800h
mov es,ax
mov si,0
mov bl,0
mov bh,23
mov cx,1920
s:
mov es:[si],bx
add si,2
loop s
mov cx,34
mov bl,45
s1:
mov es:[si],bx
add si,2
loop s1

mov cx,12
s2:
mov bl,ds:[di]
inc di
mov es:[si],bx
add si,2
loop s2

mov cx,34
mov bl,45
s3:

mov es:[si],bx
add si,2
loop s3

mov ah,4ch
int 21h

code ends
end start