实验3 转移指令跳转原理及其简单应用编程
2021-11-30 14:07 kirimi 阅读(64) 评论(5) 收藏 举报任务一
- task1.asm以及运行截图
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
- 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。
从001B跳转到000D,偏移量为-14,偏移地址为-14的补码F2
- 问题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。
从0039跳转到0029,偏移量为-16,偏移地址为-14的补码F0
任务二
- task2.asm
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
- 给出分析、调试、验证后,寄存器(ax) = ? (bx) = ? (cx) = ? 附上调试结果界面截图。
ax= 0021
bx= 0026
cx= 0772
call word将IP地址入栈,出栈时将IP存入地址赋给ax
call dword将CS:IP入栈,出栈时先付给bx IP地址,再赋给cx CS地址
任务三
- task3.asm
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 si,offset x
mov cx,len
s: mov ah,0
mov al,[si]
call printNumber
call printSpace
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
- 运行
任务四
在我的电脑上str db 'try'会出现当前CPU模式不认可的指令或寄存器错误,下面改为string
- task4.asm
assume cs:code, ds:data
data segment
string db 'try'
len equ $ - string
data ends
code segment
start:
mov ax,data
mov ds,ax
mov cx,len
mov si,offset string
mov bh,0
mov bl,2
call printStr
mov cx,len
mov si,offset string
mov bh,24
mov bl,4
call printStr
mov ah,4ch
int 21h
printStr:
mov al,0ah
mul bh
add ax,0b800h
mov es,ax
mov di,si
s: mov al,ds:[si]
mov ah,bl
mov es:[di],ax
inc si
add di,2
loop s
ret
code ends
end start
- 运行
任务五
- task5.asm
assume cs:code,ds:data
data segment
stu_no db '20498329042'
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
- 运行