实验五

  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

CPU执行任务,程序返回前,data段中的数据为多少?

  

CPU执行任务后,程序返回前,cs=076C,ss=076B,ds=076A.

程序加载后,code段的段地址为X,则data段的段地址为X-2,stack段的段地址为X-1

2

代码:

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

CPU执行程序,程序返回前,data段中的数据为多少?

data段的数据如上图

程序反悔前,cs=076C,ss=076B,ds=076A

code=x则data=x-2,stack=x-1

对于如下定义的段:

name segment

...

name ends

若数据占N个字节,则程序加载后,该段实际占有的空间为:

每个段固定16个字节,若超过16个字节,则另外再分配整数倍的段,直至全部存放下

(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

cs=076A,ss=076E,ds=076D

若code段地址为X,则data段地址为:X+3,stack段地址为:X+4

(4)没有程序入口时则ip默认为0,只有第三个程序,因为它是从代码断开始的,所以可以正确执行,别的都不是从代码断开始,所以不能

(5)

assume cs:code
 a segment
    dw 1,2,3,4,5,6,7,8,9,0ah,0bh,0ch,0dh,0eh,0fh,0ffh
a ends
b segment
    dw 0,0,0,0,0,0,0,0
b ends
code segment
start:
    mov ax,al
    mov ds,ax
    mov cx,8
    mov bx,0
    mov ax,bl
    mov ss,ax
    mov sp,10h
s:push [bx]
    add bx,2
    loop s

    mov ax,4c00h
    int 21h
code ends

 

(6)

assume cs:code
 a segment
    dw 1,2,3,4,5,6,7,8,9,0ah,0bh,0ch,0dh,0eh,0fh,0ffh
a ends
b segment
    dw 0,0,0,0,0,0,0,0
b ends
code segment
start:
    mov ax,al
    mov ds,ax
    mov cx,8
    mov bx,0
    mov ax,bl
    mov ss,ax
    mov sp,10h
   s:push [bx]
    add bx,2
    loop s

    mov ax,4c00h
    int 21h
code ends

实验与总结:有一个细节就是如果没有入口的话那ip默认为0,写程序的时候要细心一些,多写几次

posted on 2018-11-26 20:10  _刘大碗  阅读(59)  评论(0编辑  收藏  举报