8086汇编语言二重循环问题三种处理方法

1.寄存器保留CX

assume cs:code,ds:data
data segment
    db 'ibm             '  
    db 'dec             ' 
    db 'dos             '
    db 'vax             '
data ends
code segment
    start:
    mov ax,data
    mov ds,ax
    mov bx,0
    mov cx,4 
    s:mov dx,cx
    mov si,0
    mov cx,3
    s0:mov al,[bx+si]
    and al,11011111b
    mov [bx+si],al
    inc si        
    loop s0  
    mov cx,dx
    add bx,16
    loop s 
    
    mov ax,4c00h
    int 21h
code ends
end start

2.内存空间保存CX

assume cs:code,ds:data
data segment
    db 'ibm             '  
    db 'dec             ' 
    db 'dos             '
    db 'vax             '
data ends
code segment
    start:
    mov ax,data
    mov ds,ax
    mov bx,0
    mov cx,4 
    s:mov [40h],cx
    mov si,0
    mov cx,3
    s0:mov al,[bx+si]
    and al,11011111b
    mov [bx+si],al
    inc si        
    loop s0  
    mov cx,[40h]
    add bx,16
    loop s 
    
    mov ax,4c00h
    int 21h
code ends
end start

3.栈空间保存CX

assume cs:code,ds:data,ss:stack
data segment
    db 'ibm             '  
    db 'dec             ' 
    db 'dos             '
    db 'vax             '
data ends   
stack segment
    dw 0
stack ends    
code segment
    start:
    mov ax,data
    mov ds,ax  
    mov ax,stack
    mov ss,ax
    mov bx,0
    mov cx,4 
    s:push cx
    mov si,0
    mov cx,3
    s0:mov al,[bx+si]
    and al,11011111b
    mov [bx+si],al
    inc si        
    loop s0  
    pop cx
    add bx,16
    loop s 
    
    mov ax,4c00h
    int 21h
code ends
end start