When assembling this code with nasm
:
BITS 64
mov eax, 0x1
mov rax, 0x1
I get this output:
b8 01 00 00 00 b8 01 00 00 00
which is the opcode for mov eax, 0x1
repeated twice.
Does this mean that mov rax, 0x1
can always be replaced by mov eax, 0x1
or is it just in this case?
If this is correct wouldn't it be better to use than:
xor rax, rax
inc rax
as that becomes 6 bytes when assembled while mov eax, 0x1
is only 5 bytes?
Always. Most (if not all) 32-bit MOVs and ALU operations clear bits 32 through 63 of the destination register operand. See the CPU manual for details.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With