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

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 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。

14

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。

-16

2. 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

分析、调试、验证后，寄存器(ax)=0021 (bx)=0026 (cx)=076C附上调试结果界面截图。

 

 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 byte ptr bl,10
    mov si,offset x
    mov cx,len 
s:  mov al,ds:[si]
    mov ah,0
    div bl
    call printNumber
    call printSpace
    inc si
    loop s
    mov ah,4ch
    int 21h

printNumber:
    mov dx,ax
    mov ah,2
    or dl,30h
    int 21h
    mov dl,dh
    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 = $-str
data ends

stack segment
    db 16 dup(0)
stack ends

code segment
start:
    mov ax,data
    mov ds,ax
    mov ax,stack
    mov ss,ax
    mov sp,16

    mov si,offset str
    mov cx,len
    mov bl,2;绿色
    mov bh,0
    call printStr

    mov si,offset str
    mov cx,len
    mov bl,4
    mov bh,24
    call printStr

    mov ah,4ch
    int 21h

printStr:   
    mov dx,0b800h
    mov es,dx
    mov ah,0
    mov al,bh
    mov di,160
    mul di
    mov di,ax
s:  mov al,ds:[si]
    mov es:[di],al
    inc di   
    mov es:[di],bl
    inc di
    inc si
    loop s
    ret

code ends
end start
5. 实验任务5

assume cs:code,ds:data

data segment
    stu_no db '201983290198'
    len = $-stu_no
data ends

stack segment
    db 16 dup(0)
stack ends

code segment
start:
    mov ax,data
    mov ds,ax
    mov ax,stack
    mov ss,ax
    mov sp,16
    mov si,0

    call setColor
    call setLine
    call setNo

    mov ah,4ch
    int 21h

setColor:
    mov ax,0b800h
    mov es,ax
    mov ax,0
    mov al,25
    mov dl,80
    mul dl
    mov di,0
    mov bl,00010000B
    mov dl,20h 
    mov cx,ax   
s:  mov es:[di],dl
    inc di
    mov es:[di],bl
    inc di
    loop s
    ret

setLine:
    mov ax,0b800h
    mov es,ax
    mov ax,0
    mov al,24
    mov dx,160
    mul dx
    mov di,ax
    mov dl,2dh
    mov bl,00010111B
    mov cx,80
p:  mov es:[di],dl
    inc di
    mov es:[di],bl
    inc di
    loop p
    ret

setNo:
    mov ax,0b800h
    mov es,ax
    mov ax,0
    mov al,24
    mov dx,160
    mul dx
    mov di,ax
    mov ax,68
    add di,ax
    mov bl,00010111B
    mov cx,len
q:  mov ax,ds:[si]
    mov es:[di],ax
    inc di
    mov es:[di],bl
    inc di
    inc si
    loop q
    ret

code ends
end start