实验2 多个逻辑段的汇编源程序编写与调试

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

 

 

 

① 在debug中将执行到line17结束、line19之前,记录此时:寄存器(DS) = __076A__, 寄存器(SS) = _076B___, 寄存器(CS) = __076C__

② 假设程序加载后,code段的段地址是X,则,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

 

 

 

① 在debug中将执行到line17结束、line19之前,记录此时:寄存器(DS) = __076A__, 寄存器(SS) = _076B___, 寄存器(CS) = __076C__

② 假设程序加载后,code段的段地址是X,则,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

 

 

 

① 在debug中将执行到line17结束、line19之前,记录此时:寄存器(DS) = _076A___, 寄存器(SS) = _076C___, 寄存器(CS) = _076E___

② 假设程序加载后,code段的段地址是X,则,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

 

 

 

① 在debug中将执行到line9结束、line11之前,记录此时:寄存器(DS) = _076C___, 寄存器(SS) = _076E__, 寄存器(CS) = _076A___

② 假设程序加载后,code段的段地址是X,则,data段的段地址是_X+2___, stack的段地址是 _X+4___。

任务1-5

 

任务2

assume cs:code

code segment
start:
    mov ax, 0b800h
    mov ds,ax
    mov bx,0f00h
    mov cx,80
    
    mov ax, 0403h
s:  mov [bx], ax
    inc bx
    inc bx
    loop s

    mov ah, 4ch
    int 21h
code ends
end start

 

 

 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
   ; data1 ds  data2   +16B   data3 +16B
   mov bx,0
   mov dx,0
   mov cx,10
s: mov dx,0
   add dl,[bx]
   add dl,[bx+16]
   mov [bx+32],dl
   inc bx
   loop s
   
   mov ah, 4ch
   int 21h
code ends
end start

 

 

 任务4

assume cs:code

data1 segment
dw 2, 0, 4, 9, 2, 0, 1, 9 ;16B
data1 ends

data2 segment
dw 8 dup(?)
data2 ends

code segment
start: mov ax,data1
mov ds,ax
mov ax,data2
mov bx,0
mov ss,ax
mov sp,16
mov cx,8
s: push [bx]
add bx,2
loop s

mov ah, 4ch
int 21h
code ends
end start

 

 

 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

 

 

 

 

 

 源代码中line19的作用是? 将小写字母转化成大写字母

源代码中data段line4的字节数据的用途是? 设置字体的颜色

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 si, 0

    mov cx, 4
s1:
    mov dx, cx
    mov bx, 0

    mov cx, 4
s2:
    mov al, [si + bx]
    or al, 020h
    mov [si + bx], al
    inc bx
    loop s2
    mov cx, dx

    mov ax, si
    add ax, 10h
    mov si, ax
    loop s1

    mov ah, 4ch
    int 21h
code ends
end start

 

 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 cx, 5
    mov bx, 0
    mov si, 0
    mov di, 0
s:
    mov ax, ds:[bx + 2]
    mov es:[di + 2], ax
    mov ax, [bx]
    mov es:[di], ax

    mov ax, [si + 20]
    mov word ptr es:[di + 5], 0
    mov word ptr es:[di + 7], ax

    mov ax, [si + 30]
    mov word ptr es:[di + 11], ax

    mov ax, es:[di + 7]
    div byte ptr es:[di + 11]

    mov es:[di + 14], al

    add bx, 4
    add si, 2
    add di, 16
    loop s

    mov ah, 4ch
    int 21h
code ends
end start

 

 

 

posted @ 2021-11-11 18:42  08091  阅读(45)  评论(3)    收藏  举报