任务1

assume cs:code, ds:data, ss:stack
data segment
dw 0123h,0456h,0789h,0abch,0defh,0fedh,0cbah,0987h
data ends
stack segment
dw 0,0,0,0,0,0,0,0
stack ends
code segment
start: mov ax,stack
mov ss,ax
mov sp,16
mov ax,data
mov ds,ax
push ds:[0]
push ds:[2]
pop ds:[2]
pop ds:[0]
mov ax,4c00h
int 21h
code ends
end start


任务2


assume cs:code, ds:data, ss:stack
data segment
dw 0123h,0456h
data ends
stack segment
dw 0,0
stack ends
code segment
start: mov ax,stack
mov ss,ax
mov sp,16
mov ax,data
mov ds,ax
push ds:[0]
push ds:[2]
pop ds:[2]
pop ds:[0]
mov ax,4c00h
int 21h
code ends
end start

(N/16+1)*16
任务3


assume cs:code, ds:data, ss:stack
code segment
start: mov ax,stack
mov ss,ax
mov sp,16
mov ax,data
mov ds,ax
push ds:[0]
push ds:[2]
pop ds:[2]
pop ds:[0]
mov ax,4c00h
int 21h
code ends
data segment
dw 0123h,0456h
data ends
stack segment
dw 0,0
stack ends
end start

任务4

第三个程序可以执行。因为如果不指名入口,程序则从加载进内存的第一个单元起开始执行。对于前两个程序,数据段和栈段在前面,CPU会把这些数值数据当成汇编指令执行,引发错误。
任务5

assume cs:code
a segment
db 1,2,3,4,5,6,7,8
a ends
b segment
db 1,2,3,4,5,6,7,8
b ends
c segment
db 0,0,0,0,0,0,0,0
c ends
code segment
start:mov ax,a
mov ds,ax
mov ax,b
mov es,ax
mov ax,c
mov ss,ax
mov ax,0
mov bx,0
mov cx,8
s: add ax,ds:[bx]
add ax,es:[bx]
mov ss:[bx],ax
mov ax,0
add bx,1
loop s
mov ax,4c00h
int 21h
code ends
end start



任务6

assume cs:code
a segment
dw 1,2,3,4,5,6,7,8
a ends
b segment
dw 0,0,0,0,0,0,0,0
b ends
code segment
start:mov ax,a
mov ds,ax
mov ax,b
mov ss,ax
mov sp,16
mov bx,0
mov cx,8
s: push ds:[bx]
add bx,2
loop s
mov ax,4c00h
int 21h
code ends
end start


本文分析了三个8086汇编语言程序,探讨了数据段、堆栈段和代码段的使用,解释了为什么某些程序能够正确执行而其他程序则会引发错误。同时,通过具体实例讲解了不同段之间的数据交换过程。
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