understanding random number code in assembly 8086 -

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i found here on forum code generate random number between 0 , 9 in assembly 8086. here code:

    randstart:    mov ah, 2ch             int 21h           mov  ax, dx    xor  dx, dx    mov  cx, 10        div  cx           mov si, dx    mov variable, si

please, need guys explain me line:

 xor  dx, dx

i don't understand why need use xor here. hope guys can , teach me :)

xor dx,dx equals mov dx,0. logic gate that:

1+1=0  0+0=0  1+0=1  0+1=1

basically when xor number output 0. preferred on mov command because takes less memory. hope understand little better now.


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