汇编语言（七）循环结构程序设计

1.计算 1+2+3+...+n=?  其中n通过键盘输入，累加和小于2^16。

assume ds:data,cs:code
data segment
	inf1 db "Please input a number (0-65535):$"
	ibuf db 7,0,6 dup (?)
	obuf db 6 dup(?)
	inf2 db 0ah,0dh,"1+2+...+n=$"
data ends
code segment
start:	
	TenTo2 macro 
	      local again

		mov dx,offset inf1
		mov ah,09h
		int 21h
		mov dx,offset ibuf
		mov ah,0ah
		int 21h
		mov cl,ibuf+1
		mov ch,0
		mov si,offset ibuf+2
		mov ax,0
	again:	mov dx,10
		mul dx
		and byte ptr [si],0fh
		add al,[si]
		adc ah,0
		inc si
		loop again	

	endm
	
	TwoTo10 macro
	      local loop1

		mov bx,offset obuf+5
		mov byte ptr [bx],'$'
		mov cx,10
	loop1:	mov dx,0
		div cx
		add dl,30h
		dec bx
		mov [bx],dl
		or ax,ax
		jnz loop1
		mov dx,bx
		mov ah,09h
		int 21h

	endm

;;;;;;;;;;;;;;;;;;;Begin;;;;;;;;;;;;;;;;;;;;;;

	mov ax,data
	mov ds,ax
	
	TenTo2
	
	mov cx,ax
	mov ax,0
	mov bx,1
loop2:
	add ax,bx
	inc bx
	loop loop2
	
	push ax

	mov dx,offset inf2
	mov ah,09h
	int 21h

	pop ax
	TwoTo10

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

2.从自然数1开始累加，直到累加和大于60000为止，显示累加的自然数的个数和累加和。

显示格式为：1+2+...+n=sum      

其中n为累加个数，sum为累加和。

assume ds:data,cs:code
data segment

	ibuf db 7,0,6 dup (?)
	obuf db 6 dup(?)
	inf1 db 0ah,0dh,"1+2+...+$"

data ends
code segment
start:	
              TwoTo10 macro
	      local loop1

		mov bx,offset obuf+5
		mov byte ptr [bx],'$'
		mov cx,10
	loop1:	mov dx,0
		div cx
		add dl,30h
		dec bx
		mov [bx],dl
		or ax,ax
		jnz loop1
		mov dx,bx
		mov ah,09h
		int 21h

	endm

;;;;;;;;;;;;;;;;;;;Begin;;;;;;;;;;;;;;;;;;;;;;

	mov ax,data
	mov ds,ax

	mov cx,ax
	mov ax,0
	mov bx,1
loop2:
	add ax,bx
	inc bx

	cmp ax,60000
	jna loop2
	
	push bx
	push ax

        mov dx,offset inf1
	mov ah,09h
	int 21h

	dec bx
	mov ax,bx  ;>60000后 bx多加了1 故应减去
	twoto10

	mov dl,'='
	mov ah,2
	int 21h

	pop ax
	pop bx
		
	TwoTo10
	
	mov ah,4ch
	int 21h

code ends
	end start

3.从键盘输入一个6位的十进制数（<65535）,将其转化为二进制输出显示，统计该二进制数中包含1的个数，并显示统计结果。

data segment  
     infor1 db 0ah,0dh,"Please Input a Decimal Number!(<65535 && 'Esc' to exit)$"  
     infor2 db 0ah,0dh,"Your Input is Iileage!$"
     infor3 db 0ah,0dh,"Decimal To Binary is:$" 
     infor4 db 0ah,0dh,"the number of 1 is:$" 
     everbit db 0,0,0,0,0,0,0		;[5:0]存取读入的数字真实值    [6]存取1的个数

data ends  
code segment  
     assume cs:code,ds:data  
  
start:  
    print macro str  
        push ax  
        push dx  
        mov dx,offset str  
        mov ah,09h  
        int 21h  
        pop dx  
        pop ax  
    endm  
      
    prind macro ch  
        push ax  
        push dx  
        mov dl,ch  
        mov ah,02h  
        int 21h  
        pop dx  
        pop ax  
    endm  
  
    digita macro num          
        local cf,nc,ouput1  
        push ax  
	push bx
        push cx  
        push dx  
  
        mov al,num  
        mov cx,4  
        shl al,cl  
    cf:   
        shl al,1      
        jc ouput1  
        prind '0'  
    nc: loop cf   
          
        pop dx  
        pop cx 
	pop bx 
        pop ax  
  
        jmp re2  
    ouput1:   
        prind '1'  
	inc byte ptr [bx+6]
        jmp nc  

    endm  
  
;;;;;;;;;;;;;;;;;;;;;;;;begin;;;;;;;;;;;;;;;;;;;;;;;  
  
  	
        mov ax,data  
        mov ds,ax     
	mov bx,offset everbit  
          
input:        
	mov byte ptr [bx+6],0		;每次执行完计数器清0

    print infor1     
  
    mov ah,01h  
        int 21h  
    mov [bx],al  
      
    cmp al,27  
    jz exit  
  
    mov bx,offset everbit  
    mov cx,5  
    mov di,0  
save:     
   	 inc di  
  	 mov ah,01h  
       	 int 21h  
  	 mov [bx+di],al  
   	 loop save  
      
   	mov di,0  
    	mov cx,6  
        print infor3  
  
read:  
	mov al,[bx+di]  
	cmp al,byte ptr '0'  
	jb error  
	cmp al,byte ptr '9'  
	ja error  

	jmp digit

re:  
        digita al  
  
re2:        
	inc di  
        loop read  
	
	jmp prin2

    	jmp input  
  
exit:     
    	mov ah,4ch  
        int 21h  
  
error:  
        print infor2  
        jmp input  

digit:  
    	sub al,30h  
        jmp re  
prin2:
	print infor4
	mov al,[bx+6]

	cmp al,9
	jna x 
	daa
	
	push ax

	mov ah,2
	mov dl,'1'
	int 21h
	
	pop ax
	and al,0fh
	
x:	
	add al,30h

	mov dl,al
	mov ah,2
	int 21h

    	jmp input  

code ends  
     end start