Converting Decimal to Hex

问题:

First off, this is homework.

I'm trying to read a 5 digit number into the register bx. The number is assumed to be no greater than 65535 (16 bits). Below is how I am attempting to do so.

However, when I attempt to print the number, I am only printing the very last digit that was entered. Which leads me to guess that when I add another number to bx it is overwriting the previous number, but I am unable to see the problem. Any help would be appreciated, I'm almost certain that it is something small I'm overlooking :-/

mov cx,0x05 ; loop 5 times
    mov bx,0    ; clear the register we are going to store our result in
    mov dx,10   ; set our divisor to 10

read:
    mov ah,0x01     ; read a character function
    int 0x21        ; store the character in al
    sub al,0x30     ; convert ascii number to its decimal equivalent
    and ax,0x000F   ; set higher bits of ax to 0, so we are left with the decimal
    push ax         ; store the number on the stack, this is the single digit that was typed
    ; at this point we have read the char, converted it to decimal, and pushed it onto the stack
    mov ax,bx       ; move our total into ax
    mul dx          ; multiply our total by 10, to shift it right 1
    pop bx          ; pop our single digit into bx
    add bx,ax       ; add our total to bx
    loop read       ; read another char

回答1:

When using the MUL opcode, there are three different results:

• 8 bit - results are stored in ax
• 16 bit - results are stored in dx:ax
• 32 bit - results are stored in edx:eax

So when you perform your multiplication, the instruction overwrites dx with zero in your case. This means that each subsequent use of the mul opcode is multiplying by zero.