subject

Receive several one digit decimal values (0-9) from the keyboard and display their sum in the form of decimal data.

requirement:
1) The input of one decimal value is realized by subroutine
2) When the user enters directly without entering a value, the input is ended
3) The input data is multi digit decimal data, and the sum calculated inside the machine is in hexadecimal form, which needs numerical conversion, and then the result is output in string form
4) Necessary prompt information is required in the program

Example: the user enters three values at the prompt
Please input a number: 5
Please input a number: 3
Please input a number: 4
Please input a number:
The sum is:12

code

DATA SEGMENT
	STR1 DB "Please input a number: $"
	STR2 DB "The sum is: $"
	CRLF DB 0AH,0DH,'$'   ;Line feed
	COUNT DB 0
	DIVNUM DB 10
	RESULT DB 0,0,0,0,0,0,0,0
DATA ENDS
CODE SEGMENT 
ASSUME CS:CODE,DS:DATA
START:
	MOV AX,DATA
	MOV DS,AX
LOOP1:
	CALL INPUT ;Subroutine call
	CMP AL,0DH
	JZ ENDINPUT
	SUB AL,30H
	ADD COUNT,AL
	
	LEA DX,CRLF ;Line feed
	MOV AH,9
	INT 21H
	JMP LOOP1

ENDINPUT:
	LEA DX,STR2 ;Line feed
	MOV AH,9
	INT 21H
	
	MOV AL,COUNT
	MOV AH,0 ;send AX=AL As divisor
	MOV SI,0
LOOP2: 
	DIV DIVNUM   ; AL......AH
	;ADD AH,30H
	MOV RESULT[SI],AH
	CMP AL,0
	JZ NEXT
	MOV AH,0
	ADD SI,1
	JMP LOOP2

NEXT:
	ADD SI,1
	MOV CX,SI
	SUB SI,1
	
LOOP3:
	ADD RESULT[SI],30H
	MOV AH,02H
	MOV DL,RESULT[SI]
	INT 21H
	SUB SI,1
	LOOP LOOP3
	
	MOV AX,4C00H
	INT 21H
	
INPUT PROC
	LEA DX,STR1
	MOV AH,9
	INT 21H
	
	MOV AH,1
	INT 21H

	RET
INPUT ENDP	
CODE ENDS
END START

explain

1,

LOOP1:
	CALL INPUT ;Subroutine call
	CMP AL,0DH
	JZ ENDINPUT
	SUB AL,30H
	ADD COUNT,AL
	
	LEA DX,CRLF
	MOV AH,9
	INT 21H
	JMP LOOP1

LOOP, but JMP is used. Because you don't know the number of cycles, and finally jump out of the LOOP according to enter, you can not use the LOOP command (there is only one exit in this LOOP, that is, when enter is input)
Subroutine calls CALL INPUT

INPUT PROC
	LEA DX,STR1
	MOV AH,9
	INT 21H
	
	MOV AH,1
	INT 21H

	RET

The logic is simple, display the prompt STR1, and then enter the data
Register protection is not required here, because the CALL instruction is the first code, and no required registers are changed in the subroutine

CMP AL，0DH
Because the characters input in the subroutine call will be saved in AL, compare AL with the ASCII of carriage return. If carriage return, the cycle ends and the ENDINPUT jumps in

Otherwise, AL-30H will be used to convert the ASCII characters into real values, and COUNT will be used to save the cumulative sum

2,

ENDINPUT:
	LEA DX,STR2 ;Line feed
	MOV AH,9
	INT 21H
	
	MOV AL,COUNT
	MOV AH,0 ;send AX=AL As divisor
	MOV SI,0
LOOP2: 
	DIV DIVNUM   ; AL......AH
	;ADD AH,30H
	MOV RESULT[SI],AH
	CMP AL,0
	JZ NEXT
	MOV AH,0
	ADD SI,1
	JMP LOOP2

Now the real sum of all the input values has been obtained and saved in COUNT. Next, modulo 10 of COUNT one by one, each digit in COUNT in turn, and then saved in array RESULT, because modulo 10 first obtains one bit, ten bits... And finally needs to reverse output the array RESULT

	MOV AL,COUNT
	MOV AH,0 ;send AX=AL As divisor
	MOV SI,0

The divisor defaults to AX. Because COUNT is of DB type, the value of COUNT is assigned to AX. Here, SI is used to control the position where the value is stored in the RESULT array

LOOP2: 
	DIV DIVNUM   ; AL......AH
	;ADD AH,30H
	MOV RESULT[SI],AH
	CMP AL,0
	JZ NEXT
	MOV AH,0
	ADD SI,1
	JMP LOOP2

Except for 10 remainder, the real value of the number is saved in the array now (not ASCII value. 30H should be added when outputting later. I think it should be OK to add it now, but I made an error when adding it here. I simply added it at the end)
Judge whether AL is 0. If AL is 0, it proves that each bit of all COUNT is put into the array, and exit the loop. If the loop does not continue, SI+1, the next number is placed in the next bit of the array. Remember to assign AH to 0 and the divisor is AX

3,

NEXT:
	ADD SI,1
	MOV CX,SI
	SUB SI,1
	
LOOP3:
	ADD RESULT[SI],30H
	MOV AH,02H
	MOV DL,RESULT[SI]
	INT 21H
	SUB SI,1
	LOOP LOOP3

The last step is to output the number in the array

ADD SI,1
	MOV CX,SI
	SUB SI,1

Because SI is now the subscript of the last bit of the array, it needs + 1 to be the length of the array and assign CX as the number of cycles

LOOP3:
	ADD RESULT[SI],30H
	MOV AH,02H
	MOV DL,RESULT[SI]
	INT 21H
	SUB SI,1
	LOOP LOOP3

Reverse output array

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