实验3 转移指令跳转原理及其简单应用编程

1.实验任务1

程序task1.asm源码:
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

运行截图:

 

 

回答问题①
① line27, 汇编指令 loop s1 跳转时,是根据位移量跳转的。通过debug反汇编,查看其机
器码,分析其跳转的位移量是多少?(位移量数值以十进制数值回答)从CPU的角度,说明
是如何计算得到跳转后标号s1其后指令的偏移地址的。
答:line27的机器码: E2F2,所以跳转的位移量为0Dh - 1Bh = 13 - 27 = -14
 
回答问题②
② line44,汇编指令 loop s2 跳转时,是根据位移量跳转的。通过debug反汇编,查看其机
器码,分析其跳转的位移量是多少?(位移量数值以十进制数值回答)从CPU的角度,说明
是如何计算得到跳转后标号s2其后指令的偏移地址的。
答:line44机器码:E2F0,所以跳转的位移量为29h - 39h = -10h,即-16
 
问题③
③ 附上上述分析时,在debug中进行调试观察的反汇编截图 

 

 

 

 

 

2.实验任务2

程序task2.asm源码:
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 根据call指令的跳转原理,先从理论上分析,程序执行到退出(line31)之前,寄存器(ax) =  0021 寄存器(bx) = 0026 寄存器(cx) = 076C
② 对源程序进行汇编、链接,得到可执行程序task2.exe。使用debug调试,观察、验证调试结果与理论
分析结果是否一致。 

 

 结果一致

 

 

3.实验任务3

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 ax, data
    mov ds, ax
    

    mov cx,len
    mov si,0

s:  mov ah,0
    mov al,[si]
    mov bx,offset printnumber
    call bx
    mov bx,offset printSpace
    call bx
    inc si
    loop s
     

    mov ah, 4ch
    int 21h

printnumber:
    mov bl,10
    div bl
    

    mov bx,ax
    mov ah,2
    
    mov dl,bl
    or dl,30h
    int 21h
    
    mov dl,bh
    or dl,30h
    int 21h
    ret

printSpace:
    mov ah,2
    mov dl,' '
    int 21h
    ret
code ends
end start

 

 

 

4.实验任务4

assume cs:code, ds:data

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

code segment
start:  
    mov ax, data
    mov ds, ax
    mov ax,0B800H
    mov es,ax
    

    mov si,offset printStr
    mov ah,2
    mov bx,0
    call si
    
    mov si,offset printStr
    mov ah,4
    mov bx,0F00H
    call si
    
    mov ah, 4ch
    int 21h

printStr:
    mov cx,len
    mov si,0
s:  mov al,[si]
    mov es:[bx+si],ax
    inc si
    inc bx
    loop s
    ret

code ends
end start

 

 5.实验任务5

assume cs:code, ds:data

data segment
    stu_no db '20192308060' 
    len = $ - stu_no
data ends

code segment
start:  
    mov ax, data
    mov ds, ax
    mov ax,0B800H
    mov es,ax

    mov cx,0780H
    mov ah,10H
    mov al,' '
    mov bx,0

s:  mov es:[bx],ax
    add bx,2
    loop s
    

    mov cx,80
    mov ah,17H
    mov al,'-'

s1: mov es:[bx],ax
    add bx,2
    loop s1

    mov cx,len
    mov bx,0F44H
    mov si,0

s2: mov al,[si]
    mov es:[bx],ax
    inc si
    add bx,2
    loop s2

    mov ah, 4ch
    int 21h

code ends
end start

 

 

 

 

 

 

 

 
 
 
posted @ 2021-11-29 13:05  宿于江川  阅读(39)  评论(1编辑  收藏  举报