为什么我们在此语句中在 msg1 db 之后使用 0dh，0ah：msg1 db 0dh、0ah、"ENTER A CHOICE $" ;

Question

DATA segment
    msg1 db 0dh, 0ah, "ENTER A CHOICE $";
    msg2 db 0dh, 0ah, "1.Addition $";
    msg3 db 0dh, 0ah, "2.Subtraction $"; 
    msg4 db 0dh, 0ah, "3.Exit $";
    msg5 db 0dh, 0ah, "Enter the first number $";
    msg6 db 0dh, 0ah, "Enter the second number $";
    msg7 db 0dh, 0ah, "Result $";
    msg8 db 0dh, 0ah, "Enter a valid input $";
    msg9 db 0dh, 0ah, "$";
    buffer Db ?;
    ten dw 0010;
    hun dw 0100;
    tho dw 1000;
    rem dw ?;
DATA ends
CODE segment
assume CS:CODE,DS:DATA
start:
    him proc
        MOV AX,DATA; 
        MOV DS,AX; 
        MOV DX, OFFSET msg1;
        MOV AH, 09H;
        INT 21h;
        MOV DX, OFFSET msg2;
        MOV AH, 09H;
        INT 21h;
        MOV DX, OFFSET msg3;
        MOV AH, 09H;
        INT 21h;
        MOV DX, OFFSET msg4;
        MOV AH, 09H;
        INT 21h;
        MOV DX, OFFSET msg9;
        MOV AH, 09H;
        INT 21h;
        mov ah, 01h;
        int 21h;
        cmp al, 33h;
        je ex;
        cmp al, 31h;
        je addfunc;
        cmp al, 32h;
        je subfunc;
        cmp al, 33h;
        jg errfunc
        cmp al, 31h;
        jl errfunc
        proc read
            mov ah,01
            int 21h
            sub al,48
            mul ten
            mov buffer,al      
            mov ah,01           
            int 21h
            sub al,48
            add buffer,al
            ret                 ;number in buffer 
        endp
        errfunc proc
            mov dx, offset msg8
            mov ah, 09h
            int 21h
            call him
        endp
        ex proc
            MOV AH,4CH 
            INT 21H 
        endp
        addfunc proc
            MOV DX, OFFSET msg5;
            MOV AH, 09H;
            INT 21h;
            call read
            mov cl,buffer
            MOV DX, OFFSET msg6;
            MOV AH, 09H;
            INT 21h;
            call read
            mov bl,buffer
            MOV DX, OFFSET msg7;
            MOV AH, 09H;
            INT 21h;
            add cl,bl;
            mov al,cl
            mov ah,00
            call write
            call him
        endp
        subfunc proc
            MOV DX, OFFSET msg5;
            MOV AH, 09H;
            INT 21h;
            call read
            mov cl,buffer
            MOV DX, OFFSET msg6;
            MOV AH, 09H;
            INT 21h;
            call read
            mov bl,buffer
            MOV DX, OFFSET msg7;
            MOV AH, 09H;
            INT 21h;
            sub cl,bl;
            mov al,cl
            mov ah,00
            call write
            call him
        endp
        proc write;assuming value is stored in ax
            mov dx,0000
            div tho
            mov rem,dx
            add al,48
            mov dl,al
            mov ah,02h
            int 21h
            mov ax,rem
            mov dx,0000
            div hun
            mov rem,dx
            add al,48
            mov dl,al
            mov ah,02h
            int 21h
            mov ax,rem
            mov dx,0000
            div ten
            mov rem,dx
            add al,48
            mov dl,al
            mov ah,02h
            int 21h
            mov ax,rem
            mov dx,0000
            add al,48
            mov dl,al
            mov ah,02h
            int 21h
            ret         
        endp
    endp
CODE ends
end start

Answer 1

MS-DOS需要两个字符才能将光标跳到下一行的开头：0dh（Carriage Return，缩写CR）&0ah（Line Feed，缩写LF），通常缩写为CrLf。第一个字符使光标刚好跳到该行的开头，第二个字符则使光标移动到下一行。您可以使用这些定义（省略0ah、省略0dh、同时省略两者）并观察结果。

第一次使用msg1开头的CrLf不是必需的，但msg1是循环的一部分。result ...的末尾没有CrLf，所以msg的开头必须有它。