assembly - Why can't I move directly a byte to a 64 bit register?

Question

Why can't I directly move a byte from memory to a 64-bit register in Intel x86-64 assembly?

For instance, this code:

extern printf

global main

segment .text

main:
    enter   2, 0

    mov     byte [rbp - 1], 'A'
    mov     byte [rbp - 2], 'B'

    mov     r12, [rbp - 1]
    mov     r13, [rbp - 2]             

    xor     rax, rax           
    mov     rdi, Format                                                                                             
    mov     rsi, r12                                                                                                
    mov     rdx, r13                                                                                                
    call    printf                                                                                                  

    leave                                                                                                           
    ret                                                                                                             

segment .data                                                                                                       
Format:     db "%d %d", 10, 0

prints:

65 16706

I need to change the move byte to registers r12 and r13 to this in order to make the code work properly:

xor     rax, rax
mov     al, byte [rbp - 1]
mov     r12, rax
xor     rax, rax
mov     al, byte [rbp - 2]
mov     r13, rax

Now, it prints what is intended:

65 66

Why do we need to do this?

Is there a simpler way of doing this?

Thanks.

Answer 1

Use move with zero or sign extension as appropriate.

For example: movzx eax, byte [rbp - 1] to zero-extend into RAX.

movsx rax, byte [rbp - 1] to sign-extend into RAX.