汇编试验六：实践课程中的程序

（1）大小写转换

Source Code:

assume cs:codesg,ds:datasg

datasg segment
    db 'BaSiC'
    db 'iNfOrMaTiOn'
datasg ends

codesg segment

start:
    mov ax,datasg
    mov ds,ax

    mov bx,0
    mov cx,5

s:  mov al,[bx]
    and al,11011111B
    mov [bx],al
    inc bx
    loop s


    mov bx,5
    mov cx,11
s0: mov al,[bx]
    or al,00100000B
    mov [bx],al
    inc bx
    loop s0

    mov ax,4c00H
    int 21H

codesg ends
end start

效果如图：

小写转大写and操作，大写转小写or操作

（2）两个字符串等长，则可以用数组的形式访问这两个字符串

Source Code:

assume cs:codesg , ds:datasg

datasg segment
    db 'BaSiC'
    db 'MinIX'
datasg ends

codesg segment

start:
    mov ax,datasg
    mov ds,ax

    mov bx,0
    mov cx,5

s:
    mov al,[bx]
    and al,11011111b
    mov [bx],al

    mov al,[5+bx]
    or al,00100000b
    mov [5+bx],al
    inc bx
    loop s

    mov ax,4c00H
    int 21H

codesg ends
end start

效果如下：

（也就是[bx+idata]形式）

 或者0[bx]，同[bx+idata]

Source Code:

assume cs:codesg,ds:datasg

datasg segment
    db 'BaSiC'
    db 'MinIX'
datasg ends

codesg segment
start:
    mov ax,datasg
    mov ds,ax

    mov bx,0
    mov cx,5

s:
    mov al,0[bx]
    and al,11011111b
    mov 0[bx],al
    mov al,5[bx]
    or al,00100000b

    mov 5[bx],al
    inc bx
    loop s

    mov ax,4c00H
    int 21H

codesg ends
end start

（3）SI  DI 两个下标寄存器，复制字符串；

（一个字一个字复制）

Source Code:

assume cs:codesg ,ds:datasg

datasg segment
    db 'welcome to masm!'
    db '................'
datasg ends

codesg segment
start:
    mov ax,datasg
    mov ds,ax
    mov si,0
    mov di,16
    mov cx,8
s:
    mov ax,[si]
    mov [di],ax
    add si,2
    add di,2
    loop s

    mov ax,4c00H
    int 21H
codesg ends
end start

（4）双重循环，将cx保存到dx中；

Source Code:

assume cs:codesg,ds:datasg

datasg segment
    db 'ibm             '
    db 'dec             '
    db 'dos             '
    db 'vax             '
datasg ends

codesg segment
start:
    mov ax,datasg
    mov ds,ax
    mov bx,0

    mov cx,4

    s0:
        mov dx,cx
        mov si,0
        mov cx,3

    s:
        mov al,[bx+si]
        and al,11011111b
        mov [bx+si],al
        inc si
        loop s

        add bx,16
        mov cx,dx
        loop s0

    mov ax,4c00H
    int 21H

codesg ends
end start

效果如图：

（5）数据的暂存，寄存器个数少，一般思路是放到栈里；

Source Code:

assume cs:codesg, ds:datasg, ss:stacksg

datasg segment
    db 'ibm             '
    db 'dec             '
    db 'dos             '
    db 'vax             '
datasg ends

stacksg segment
    dw 0,0,0,0,0,0,0,0
stacksg ends

codesg segment

start:
    mov ax,stacksg
    mov ss,ax
    mov sp,16

    mov ax,datasg
    mov ds,ax
    mov bx,0

    mov cx,4

    s0:
        push cx
        mov si,0
        mov cx,3

        s:
            mov al,[bx+si]
            and al,11011111b
            mov [bx+si],al
            inc si
            loop s

        add bx,16
        pop cx
        loop s0

    mov ax,4c00H
    int 21H

codesg ends
end start

效果如下：

（6）问题7.9，只需将前面的bx起始位置修改；

效果如下：

Source Code:

assume cs:codesg, ds:datasg, ss:stacksg

stacksg segment
    dw 0,0,0,0,0,0,0,0
stacksg ends

datasg segment
    db '1. display      '
    db '2. brows        '
    db '3. replace      '
    db '4. modify       '
datasg ends

stacksg segment
    dw 0,0,0,0,0,0,0,0
stacksg ends

codesg segment

start:
    mov ax,stacksg
    mov ss,ax
    mov sp,16

    mov ax,datasg
    mov ds,ax
    mov bx,3

    mov cx,4

    s0:
        push cx
        mov si,0
        mov cx,3

        s:
            mov al,[bx+si]
            and al,11011111b
            mov [bx+si],al
            inc si
            loop s

        add bx,16
        pop cx
        loop s0

    mov ax,4c00H
    int 21H

codesg ends
end start