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