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
