Assembling language realizes cycle programming experiment

Purpose:

1. Master the use of comparison instructions, transfer instructions, and loop instructions.

2. Further master the usage of debugging tools.

Experiment content:

Calculate 1+2+.....+n=?, where n is entered through the keyboard, and the cumulative sum is less than 2^16. The expected result is as follows:

Program flow chart:

 Program code (with explanation):

Program points:

1. Receive the decimal number input by the keyboard and convert it into a binary number (stored in the AX register) instruction sequence

2. How to display the entered number as the target image in the print result

3. Convert the binary number stored in AX to decimal number and display the instruction sequence

;数据段

DATA	SEGMENT
	INF1  DB "Please input a number(1-361):$"    ;$处截止
	INF2  DB "1+2+...+$"
	INF3  DB "=$"            ;为了实现输出结果设置两个字符串进行连接评凑结果 
	IBUF  DB 7,0,6 DUP(?)
    ;存放输入的值
    ;注意IBUF+1为输入数的位数，IBUF+2以后的及存放的值，比如：360 按33H,36H,30H存放
	OBUF  DB 6 DUP(?)         ;输出结果
DATA	ENDS
;代码段

CODE	SEGMENT
	ASSUME CS:CODE,DS:DATA
START:	
	MOV  AX,DATA
	MOV  DS,AX

	MOV  DX,OFFSET INF1    ;输出INF1字符串，值存入IBUF1中，利用9号功能键
	MOV  AH,09H        
	INT  21H

	MOV  DX,OFFSET IBUF    ;在此键入十进制数
	MOV  AH,0AH
	INT  21H

	MOV  CL,IBUF+1        ;十进制数的位数送给计数器进行循环
	MOV  CH,0
	MOV  SI,OFFSET IBUF+2    ;取偏移地址，找到存第一个数值的位置，给SI，若为360，这时[SI]为3
	MOV  AX,0            ;开始将十进制转为二进制
AGAIN:	
	MOV  DX,10            ;((0*10+a4)*10+...)*10+a0
	MUL  DX
	AND  BYTE PTR [SI],0FH      ;将相应存储单元中的字符ASCII转换为数字
	ADD  AL,[SI]        ;2条指令的功能为(AX)<=(AX)+数字
	ADC  AH,0            
	INC  SI            ;[SI]指向下一个
	LOOP AGAIN        ;循环执行，完成(AX)<=(AX)*10+[SI]
	

	MOV  CX,AX
	MOV  AX,0
	MOV  BX,1
LOOP2:  
	ADD  AX,BX        ;这里分析流程图吧
	INC  BX
	LOOP LOOP2
	

	MOV  BX,OFFSET OBUF+5        ;开始二进制转为10进制
	MOV  BYTE PTR [BX],'$'
	
	MOV  CX,10        ;作（DX）:(AX)/10运算
LOOP1:  
	MOV  DX,0        ;被除数高16位清零
	DIV  CX
	ADD  DL,30H        ;将10进制数转换为ASCII码
	DEC  BX    
	MOV  [BX],DL
	OR   AX,AX
	JNZ  LOOP1
                    ;判断商是否为0，不为0继续


	MOV  DL,0AH        ;打印回车换行符，利用2号功能键
	MOV  AH,02H
	INT  21H
	
	MOV  DX,OFFSET INF2    ;打印INF2字符串
	MOV  AH,09H
	INT  21H
	
	MOV  CL,IBUF+1        ;实现将输入的值打印出来，符合格式要求
	MOV  SI,OFFSET IBUF+2
LOOP3:  
	MOV  DL,[SI]        ;逐个字符打印输入的值n
    ADC  DL,30H
    MOV  AH,02H
    INT  21H
    
    INC  SI
    DEC  CL
    JNZ  LOOP3
	
	MOV  DX,OFFSET INF3    ;打印INF3字符串，凑格式
	MOV  AH,09H
	INT  21H	

	MOV  DX,BX        ;显示求和结果转换得到的十进制数
	MOV  AH,09H
	INT  21H

	MOV  AH,4CH
	INT  21H
CODE	ENDS
	END  START

Realize the result:

It is found that there is no problem in entering two or three digits