实验2 多个逻辑段的汇编源程序编写与调试
2021/11/6 11:40:23
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1.实验任务1
任务1-1
assume ds:data,cs:code,ss:stack data segment db 16 dup(0) 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 ah,4ch int 21h code ends end start
寄存器(DS) = 076a,寄存器(SS) = 076b,寄存器(CS) = 076c
data段的段地址是 X-2,stack的段地址是 X-1
任务1-2
assume ds:data,cs:code,ss:stack data segment db 4 dup(0) data ends stack segment db 8 dup(0) stack ends code segment start: mov ax,data mov ds,ax mov ax,stack mov ss,ax mov sp,8 mov ah,4ch int 21h code ends end start
寄存器(DS) = 076a,寄存器(SS) = 076b,寄存器(CS) = 076c
data段的段地址是 X-2,stack的段地址是 X-1
任务1-3
assume ds:data, cs:code, ss:stack data segment db 20 dup(0) data ends stack segment db 20 dup(0) stack ends code segment start: mov ax, data mov ds, ax mov ax, stack mov ss, ax mov sp, 20 mov ah, 4ch int 21h code ends end start
寄存器(DS) = 076a,寄存器(SS) = 076c,寄存器(CS) = 076e
data段的段地址是 X-4,stack的段地址是 X-2
任务1-4
assume ds:data, cs:code, ss:stack code segment start: mov ax, data mov ds, ax mov ax, stack mov ss, ax mov sp, 20 mov ah, 4ch int 21h code ends data segment db 20 dup(0) data ends stack segment db 20 dup(0) stack ends end start
寄存器(DS) = 076c,寄存器(SS) = 076e,寄存器(CS) = 076a
data段的段地址是 X+2,stack的段地址是 X+4
任务1-5
①((N+15)/16)*16
②task1_4.asm可以正确执行。因为如果不指明程序的入口,程序将从我们代码的第一行开始读取。其余task的第一行是我们定义的数据段,不是代码段,因此没有相对应的指令,程序执行时将会发生错误。
2.实验任务2
assume cs:code code segment start: mov ax,0b800h mov ds,ax mov bx,0f00h mov cx,80 s: mov [bx],0403h add bx,2 loop s mov ax,4c00h int 21h code ends end start
3.实验任务3
assume cs:code data1 segment db 50, 48, 50, 50, 0, 48, 49, 0, 48, 49 ; ten numbers data1 ends data2 segment db 0, 0, 0, 0, 47, 0, 0, 47, 0, 0 ; ten numbers data2 ends data3 segment db 16 dup(0) data3 ends code segment start: mov ax,data1 mov ds,ax mov bx,0 mov cx,10 s: mov ax,[bx] add ax,[bx].16 mov [bx].32,ax inc bx loop s mov ah,4ch int 21h code ends end start
![](/upload/202111/06/202111061140194722.png)
4.实验任务4
assume cs:code data1 segment dw 2, 0, 4, 9, 2, 0, 1, 9 data1 ends data2 segment dw 8 dup(?) data2 ends code segment start: mov ax,data1 mov ds,ax mov bx,0 mov si,30 mov cx,8 s: mov ax,[bx] mov [si],ax sub si,2 add bx,2 loop s mov ah, 4ch int 21h code ends end start
![](/upload/202111/06/202111061140204512.png)
5.实验任务5
assume cs:code, ds:data data segment db 'Nuist' db 2, 3, 4, 5, 6 data ends code segment start: mov ax, data mov ds, ax mov ax, 0b800H mov es, ax mov cx, 5 mov si, 0 mov di, 0f00h s: mov al, [si] and al, 0dfh mov es:[di], al mov al, [5+si] mov es:[di+1], al inc si add di, 2 loop s mov ah, 4ch int 21h code ends end start
![](/upload/202111/06/202111061140211773.png)
line19的作用是:
0dfh转换为2进制就是11011111,将其与al进行and操作后,第5位(从0开始)变成0,相当于ascii码减去32,也就是把字母转换为大写字母。
line4的作用是:
用途是设置字符的颜色。
6.实验任务6
assume cs:code, ds:data data segment db 'Pink Floyd ' db 'JOAN Baez ' db 'NEIL Young ' db 'Joan Lennon ' data ends code segment start: mov ax,data mov ds,ax mov bx,0 mov cx,4 s: or byte ptr [bx],32 add bx,16 loop s mov ah, 4ch int 21h code ends end start
![](/upload/202111/06/202111061140220347.png)
7.实验任务7
assume cs:code, ds:data, es:table data segment db '1975', '1976', '1977', '1978', '1979' dw 16, 22, 382, 1356, 2390 dw 3, 7, 9, 13, 28 data ends table segment db 5 dup( 16 dup(' ') ) ; table ends code segment start: mov ax,data mov ds,ax mov ax,table mov es,ax mov bx,0 mov bp,0 mov si,20 mov cx,5 s: mov ax,ds:[bx] mov es:[bp],ax mov ax,ds:[bx+2] mov es:[bp+2],ax mov ax,ds:[si] mov es:[bp+5],ax mov word ptr es:[bp+7],0 mov ax,ds:[si+10] mov es:[bp+10],ax mov ax,ds:[si] mov dl,ds:[si+10] div dl mov es:[bp+13],al mov byte ptr es:[bp+14],0 add bx,4 add bp,16 add si,2 loop s mov ah, 4ch int 21h code ends end start
![](/upload/202111/06/202111061140228423.png)
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