How can I move an 8-bit address into a 16-bit register in x86 assembly?

Question

Here, I\'m trying to move the variable X (which is an 8-bit variable) into the register bx (which is a 16-bit register). How can I move the value of X into the register bx i

Answer 1

In addition to Rahul's answer, if you also need to zero out bh and are working on anything 80386 or newer (as indicated by the .686p) is:

movzx bx, X

Of if you are using X as a signed value and needs to sign-extend bx:

movsx bx, X

Answer 2

The low 8-bits of BX are addressable as BL.

So, all you need to do is: mov bl, X