实验3 转移指令跳转原理及其简单应用编程

1.实验任务1

task1.asm源码：

assume cs:code, ds:data

data segment
    x db 1, 9, 3
    len1 equ $ - x

    y dw 1, 9, 3
    len2 equ $ - y
data ends

code segment
start:
    mov ax, data
    mov ds, ax

    mov si, offset x
    mov cx, len1
    mov ah, 2
 s1:mov dl, [si]
    or dl, 30h
    int 21h

    mov dl, ' '
    int 21h

    inc si
    loop s1

    mov ah, 2
    mov dl, 0ah
    int 21h

    mov si, offset y
    mov cx, len2/2
    mov ah, 2
 s2:mov dx, [si]
    or dl, 30h
    int 21h

    mov dl, ' '
    int 21h

    add si, 2
    loop s2

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

运行截图：

问题一：

跳转的位移量为-14(F2的补码)，CPU通过loop指令下一条指令的地址（loop指令执行完需要自加）减去s1所在地址得到偏移地址。

 

问题二：

跳转的位移量为-16(F0的补码)，CPU通过loop指令下一条指令的地址（loop指令执行完需要自加）减去s2所在地址得到偏移地址。

 

2.实验任务2

task2.asm源码：

assume cs:code, ds:data

data segment
    dw 200h, 0h, 230h, 0h
data ends

stack segment
    db 16 dup(0)
stack ends

code segment
start:  
    mov ax, data
    mov ds, ax

    mov word ptr ds:[0], offset s1
    mov word ptr ds:[2], offset s2
    mov ds:[4], cs

    mov ax, stack
    mov ss, ax
    mov sp, 16

    call word ptr ds:[0]
s1: pop ax

    call dword ptr ds:[2]
s2: pop bx
    pop cx

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

问题一：

寄存器(ax) =pop ax指令的地址，寄存器(bx) =pop bx指令的地址， 寄存器(cx) =cs中存的地址

问题二：

和理论上一致，调试结果如下：

3.实验任务3

task3源码：

assume cs:code, ds:data

data segment
    x db 99, 72, 85, 63, 89, 97, 55
    len equ $- x
data ends

code segment
start:
    mov ax,data
    mov ds,ax
    mov cx,len
    mov bx,0
    mov byte ptr ds:[len],10

s:    mov ax,0
    mov al,ds:[bx]
    call printNumber
    call printSpace
    inc bx
    loop s

    mov ah,4ch
    int 21h

printNumber:
    div byte ptr ds:[len]
    mov dx,ax
    or dl,30h
    mov ah,2
    int 21h

    mov dl,dh
    or dl,30h
    mov ah,2
    int 21h
    
    ret

printSpace:
    mov dl,' '
    mov ah,2
    int 21h
    ret

code ends
end start

运行截图：

4.实验任务4

task4源码：

assume cs:code, ds:data

data segment
    str db 'try'
    len equ $ - str
data ends

code segment
start:
    mov ax,data
    mov ds,ax

    mov si,0
    mov cx,len
    mov bl,00000010b
    mov bh,0
    call printStr

    mov si,0
    mov cx,len
    mov bl,00000100b
    mov bh,24
    call printStr

    mov ah,4ch
    int 21h

printStr:
    mov ax,0b800h
    mov es,ax
    mov ax,0
    mov al,bh
    mov dx,160
    mul dx
    mov di,ax
    s:    mov al,ds:[si]
        mov es:[di],al
        inc si
        inc di
        mov es:[di],bl
        inc di
        loop s
    ret
code ends
end start

运行截图：

5.实验任务5

task5源码：

assume cs:code, ds:data

data segment
    stu_no db '201983290123'
    len = $ - stu_no
    len1 = (80-len)/2
data ends

code segment
start:
    mov ax,data
    mov ds,ax
    
    mov bl,00010000b
    call printColor

    mov si,0
    mov bl,00010111b
    mov bh,24
    call printStr
    mov ah,4ch
    int 21h

printColor:
    mov ax,0b800h
    mov es,ax
    mov ax,0
    mov al,25
    mov dx,0
    mov dx,80
    mul dx
    mov cx,ax
    mov al,' '
    mov di,0
    s:    mov es:[di],al
        inc di
        mov es:[di],bl
        inc di
        loop s
    ret

printStr:
    mov ax,0b800h
    mov es,ax
    mov ax,0
    mov al,bh
    mov dx,160
    mul dx
    mov di,ax

    mov al,'-'
    mov cx,len1
    s1:    mov es:[di],al
        inc di
        mov es:[di],bl
        inc di
        loop s1
    
    mov cx,len
    s2:    mov al,ds:[si]
        mov es:[di],al
        inc si
        inc di
        mov es:[di],bl
        inc di
        loop s2

    mov al,'-'
    mov cx,len1
    s3:    mov es:[di],al
        inc di
        mov es:[di],bl
        inc di
        loop s3
    ret
code ends
end start

运行截图：