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汇编实验：

 2 years ago
source link: https://seineo.github.io/%E6%B1%87%E7%BC%96%E5%AE%9E%E9%AA%8C%EF%BC%9A%E7%BB%9F%E8%AE%A1%E6%95%B0%E5%AD%97%E4%B8%AA%E6%95%B0.html
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输入一个字符串，统计数字出现的次数。

算法流程图

data segment
buffer db 5 dup(?)
input db 'please input a number <= 5 digits', 0ah, 0dh, '$'
outOfBound db 0ah, 0dh, 'error: more than 5 digits',0ah, 0dh, '$'
notANum db 0ah, 0dh, 'error: not a number',0ah, 0dh,'$'
data ends

code segment
	      assume ds:data, cs:code
	start:
		mov bl, 0
	l4:
		mov ah, 01h
		int 21h
		cmp al, 20h   ; 空格结束输入
		je count
		cmp al, byte ptr '0'
		jb l4
		cmp al, byte ptr '9'
		ja l4
		inc bl
		jmp l4
	count:    
   		mov ah, 0
    	mov al, bl    ; 将要输出的数字
    	or  ax, ax
    	jz  zero    ; 是零则直接输出
    	mov bx, -1    ; 栈底
    	push bx
    	mov bx, 10    ; 除数
	l5:
		xor dx, dx   ; 余数清零
   		div bx
   		mov cx , ax   ;商
   		or  cx , dx
   		jz  print   ; 商与余数全零则结束
   		push dx
   		jmp l5
	zero:  
		mov dl , 30h
   		mov ah , 02h
   		int 21h
		jmp exit
   
	print:
   		pop dx
   		cmp dx , -1
   		je  exit
   		add dx , 30h
   		mov ah , 02h
   		int 21h
   		jmp print

	exit:
		mov ah, 08h  ;结束时停留
		int 21h
  	 	mov ah , 4ch                  ;返回DOS
  	 	int 21h
code ends
end start

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