Why can't I move directly a byte to a 64 bit register?
assembly, x86-64
Solution
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.
Problem
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.