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式。
但是,当我尝试打印数字时,我只打印了输入的最后一个数字。这让我猜测,当我向 bx 添加另一个数字时,它会覆盖前一个数字,但我看不到问题所在。任何帮助将不胜感激,我几乎可以肯定这是我忽略的小事:-/
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
最佳答案
使用 MUL 操作码时,有三种不同的结果:
edx:eax
因此,当您执行乘法时,在您的情况下,指令会用零覆盖 dx。这意味着 mul 操作码的每次后续使用都乘以零。
关于assembly - 将十进制转换为十六进制,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/5533119/