实验一：
(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
cs:076c ds: 076a ss:076b
data段的段地址是X-2， stack的段地址是X-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
cs:076c ds: 076a ss:076b
data段的段地址是X-2， stack的段地址是X-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
cs:076e ds: 076a ss:076c
data段的段地址是X-4， stack的段地址是X-2
(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
cs:076a ds: 076c ss:076e
data段的段地址是X+2， stack的段地址是X+4
(5)
1）实际分配的内存为((N/16+1)*16
2）task1_4.asm仍然可以执行；因为“end start”改为“end”，则程序不能识别 start，会从程序开头，数据段和栈段开始执行，不是从程序代码开始，数据段和栈段所代表的代码则是乱码
实验二：
assume cs:code
code segment
mov ax, 0b800h
mov ds, ax
mov bx, 0f00h
mov ax, 0403h
mov cx, 160
s: mov [bx], ax
add bx, 2
loop s
mov ah, 4ch
int 21h
code ends
end
结果：
实验三：
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 bx,0 ;
mov dx,0 ;
mov cx,10 ;
s:
mov dx,0
mov ax,data1
mov ds,ax
add dl,[bx]
mov ax,data2
mov ds,ax
add dl,[bx]
mov ax,data3
mov ds,ax
mov [bx],dl
inc bx
loop s
code ends
end start
相加前
data1：
data2：
data3：
反汇编：
结果：
实验四：
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 ax,data2
mov ss,ax
mov sp,16
mov bx,0
mov cx,8
s:
push ds:[bx]
add bx,2
loop s
mov ah, 4ch
int 21h
code ends
end start
在076A：0010处显示的是data2中的内容
实验五：
运行结果：
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
(2)
（3)line25执行之后、line27执行之前，观察 结果：
（4）源代码中line19的作用是：把单片机的AL与DF(1101 1111)进行与，就是把左边数过去的第三位清0
（5）line4中的字节数据，用于控制5个字符每个字符的颜色
实验六：
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, 5
s:
mov al, [bx]
or al, 00100000B
mov [bx], al
add bx, 16
loop s
mov ah, 4ch
int 21h
code ends
end start
调试：
反汇编：
结果：
实验七：
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 si,0
mov di,0
mov cx,5
s:
mov ax,ds:[si]
mov es:[di], ax
mov ax,ds:[si+2]
mov es:[di+2], ax
mov ax,ds:[si+84]
mov es:[di+5],ax
mov dx,ds:[si+84+2]
mov es:[di+7],dx
push cx
mov cx,ds:[84+84+bx]
mov es:[di+0ah],cx
div cx
pop cx
mov es:[di+0dh],ax
add si,4
add bx,2
add di,16
loop s
mov ah, 4ch
int 21h
code ends
end start
table原始数据：
运行完后：