实验3 转移指令跳转原理及其简单应用编程

实验任务1

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,line27的机器码: E2F2，所以跳转的位移量为0Dh - 1Bh = 13 - 27 = -14

 2,line44机器码:E2F0，所以跳转的位移量为29h - 39h = -10h，即-16

3，

 

 

 实验任务2

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) =  s1的偏移地址       寄存器(bx) = s2 的偏移地址      寄存器(cx) = cs中存储的值

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。 

 ax为0021是s1的偏移地址

bx为0026是s2的偏移地址

cx为076c为cs中的值

 所以结果一致

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

　　

 

 

 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

　　

 

 

 

 实验任务5

assume cs:code, ds:data

data segment
    stu_no db '201983290306' 
    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