How can I load a single byte from address? I thought it would be something like this:
mov rax, byte[rdi]
-
A duplicate with more detailed answers: Why can't I move directly a byte to a 64 bit register?Peter Cordes– Peter Cordes2021-12-16 23:26:20 +00:00Commented Dec 16, 2021 at 23:26
Add a comment
|
1 Answer
mov al, [rdi]
Merge a byte into the low byte of RAX.
Or better, avoid a false dependency on the old value of RAX by zero-extending into a 32-bit register (and thus implicitly to 64 bits) with MOVZX:
movzx eax, byte [rdi] ; most efficient way to load one byte on modern x86
Or if you want sign-extension into a wider register, use MOVSX.
movsx eax, byte [rdi] ; sign extend to 32-bit, zero-extend to 64
movsx rax, byte [rdi] ; sign extend to 64-bit
(On some CPUs, MOVSX is just as efficient as MOVZX, handled right in a load port without even needing an ALU uop. https://uops.info. But there are some where MOVZX loads are cheaper than MOVSX, so prefer MOVZX if you don't care about the upper bytes and really just want to avoid partial-register shenanigans.)
The MASM equivalent replaces byte with byte ptr.
A mov load doesn't need a size specifier (al destination implies byte operand-size). movzx always does for a memory source because a 32-bit destination doesn't disambiguate between 8 vs. 16-bit memory sources.
The AT&T equivalent is movzbl (%rdi), %eax (with movzb specifying that we zero-extend a byte, the l specifying 32-bit destination size.)
