Question

我需要提示用户一个告诉他写一个号码的msg，然后我存储这个号码并对其进行一些操作在INT 21h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” 我将其视为3,5,7

这不是我的目标。任何想法？

Answer 1

When you managed to get the user input，将其指针放在ESI（ESI =地址到字符串）

中

.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

Answer 2

获得字符串后，必须将其转换为数字。问题是，你必须编写自己的程序来做到这一点。这是我经常使用的（虽然用C语写）：

int strToNum(char *s) {
    int len = strlen(s), res = 0, mul = 0;
    char *ptr = s + len;

    while(ptr >= s)
        res += (*ptr-- - '0') * (int)pow(10.0, mul++);

    return res;
}

这是解释。首先，*ptr-- - '0'获取数字的整数表示（以便'9' - '0' = 9，然后它递减ptr，以便它指向前一个字符。一旦我们知道该数字，我们就有了将其提高到10的幂。例如，假设输入为'357'，代码的作用是：

('7' - '0' = 7) * 10 ^ 0 =   7 +
('5' - '0' = 5) * 10 ^ 1 =  50 +
('3' - '0' = 3) * 10 ^ 2 = 300 = 
---------------------------------
                           357

使用INT 21h（DOS）＆amp;读取数字8086组装

2 个答案: