实验2 多个逻辑段的汇编源程序编写与调试

任务1

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 ds：076A   ss：076B cs：076C

2 x-2，x-1

 

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 ds:076A     ss:076B   cs:076C

2 data:x-2             stack:x-1

 

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 ds:076A     ss:076C   cs:076E

2 data:x-2             stack:x-1

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 ds:076C     ss:076E   cs:076A

2 data:x+2             stack:x+4

1-5

1 (N/16+1)H

2 end指程序结束

 End start 指程序结束，第一条指令在start处，所以task1_5可以运行，因为start恰好就是第一条指令

 

任务2

assume cs:code
code segment
mov ax,0b800h
mov ds,ax
mov ax,0403h
mov bx,0f00h
mov cx,80
s: mov [bx],ax
   inc bx
   inc bx
   loop s
mov ax,4c00h
int 21h
code ends
end

 

 

 

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

 

 

 

 

 

 

 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,16
mov bp,16
s:
mov ax,[bp]
mov [bx],ax
add bx,2
sub bp,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

 

 

 

 

 

 

进行相与运算，结果-32，即把大写字母转化为小写字母

数值的作用是用来控制输出的颜色

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 bx, 0
mov cx, 4
s: 
mov ax, [bx]
or ax, 20h
mov [bx], ax
add bx, 16
loop s

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 bx, 0
    mov bp, 0
    mov cx, 5
years:
    mov ax, ds:[bx]
    mov es:[bp], ax
    mov ax, ds:[bx+2]
    mov es:[bp+2], ax
    add bx, 4
    add bp, 10h
loop years

    mov bp, 5
    mov cx, 5
income:
    mov ax, ds:[bx]
    mov es:[bp], ax
    add bx, 2
    add bp, 10h
loop income

    mov cx, 5
    mov bp, 10
people:
    mov ax, ds:[bx]
    mov es:[bp], ax
    add bx, 2
    add bp, 10h
loop people

    mov cx, 5
    mov bp, 5
average:
    mov ax, es:[bp]
    mov bl, es:[bp+5]
    div bl
    mov es:[bp+8], al
    add bp,10h
loop average

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