Reading a number using INT 21h (DOS) & 8086 assmebly

前端 未结 2 1965
借酒劲吻你
借酒劲吻你 2020-12-06 21:06

I need to prompt to user a msg that tells him to write a number , then I store this number and do some operation on it After searching in INT 21h I found this :

         


        
相关标签:
2条回答
  • 2020-12-06 21:43

    When you managed to get the user input, put the its pointer in ESI (ESI = address to the string)

    .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
    
    0 讨论(0)
  • 2020-12-06 21:49

    Once you've got the string you have to convert it to number. The problem is, you have to code your own procedure to do that. This is the one I usually use (written in C though):

    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;
    }
    

    Here's the explanation. First of all, *ptr-- - '0' gets the integer representation of a number (so that '9' - '0' = 9, then it decremenst ptr so that it points to the previous char. Once we know that number, we have to raise it to a power of 10. For example, suppose the input is '357', what the code does is:

    ('7' - '0' = 7) * 10 ^ 0 =   7 +
    ('5' - '0' = 5) * 10 ^ 1 =  50 +
    ('3' - '0' = 3) * 10 ^ 2 = 300 = 
    ---------------------------------
                               357
    
    0 讨论(0)
提交回复
热议问题