assembly - 使用 INT 21h (DOS) 和 8086 汇编读取数字

Question

我需要提示用户一条消息，告诉他写一个数字，然后我存储这个数字并对其进行一些操作在 INT 21h 中搜索后我发现了这个:

INT 21h / AH=1 - read character from standard input, with echo, result is stored in AL.
if there is no character in the keyboard buffer, the function waits until any key is pressed. 

example:

    mov ah, 1
    int 21h

主要问题是它只能读取一个字符并将其表示为 ASCII所以如果我需要写数字“357”我将其读作 3 , 5 , 7

这不是我的目标。有什么想法吗？

Answer 1

When you managed to get the user input ，将其指针放入 ESI(ESI = 字符串的地址)

.DATA
myNumber BYTE "12345",0        ;for test purpose I declare a string '12345'

Main Proc
    xor ebx,ebx                ;EBX = 0
    mov  esi,offset myNumber   ;ESI points to '12345'

loopme:

    lodsb                      ;load the first byte pointed by ESI in al

    cmp al,'0'                 ;check if it's an ascii number [0-9]
    jb noascii                 ;not ascii, exit
    cmp al,'9'                 ;check the if it's an ascii number [0-9]
    ja noascii                 ;not ascii, exit

    sub al,30h                 ;ascii '0' = 30h, ascii '1' = 31h ...etc.
    cbw                        ;byte to word
    cwd                        ;word to dword
    push eax
    mov eax,ebx                ;EBX will contain '12345' in hexadecimal
    mov ecx,10
    mul ecx                    ;AX=AX*10
    mov ebx,eax
    pop eax
    add ebx,eax
    jmp loopme                 ;continue until ESI points to a non-ascii [0-9] character
    noascii:
    ret                        ;EBX = 0x00003039 = 12345
Main EndP