将十进制转换为十六进制 [英] Converting Decimal to Hex

问题描述

首先，这是家庭作业.

我正在尝试将5位数字读入寄存器bx.假定该数字不大于65535(16位).以下是我尝试这样做的方法.

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.

但是，当我尝试打印数字时，我仅打印输入的最后一位数字.这使我猜测，当我向bx添加另一个数字时，它会覆盖先前的数字，但是我看不到问题所在.任何帮助将不胜感激，我几乎可以肯定它是我忽略的小东西:-/

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

推荐答案

使用MUL操作码时，会有三种不同的结果:

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

• 8位-结果存储在斧头中
• 16位-结果存储在dx:ax中
• 32位-结果存储在 edx:eax

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

因此，当您执行乘法运算时，在您的情况下，指令将dx覆盖为零.这意味着对mul操作码的每次后续使用都乘以零.

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.

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