How to swap bytes in 64bit Registers using NASM?

Question

I've been searching for an equivalent bswap instruction for 64 bit, but I did not find anything yet.

Let's say I have the following in a register RAX:

mov rax, 0x666E69

After swaping the values the register RAX should contain the following:

rax = 0x696E6600

I already tried with shl, shr, rol, and ror but nothing worked. Can you please help me ? Thank you.

If you wanted to swap 64 bits I would expect input 666E69h to result in 696E6600_00000000h. Which one is it? — ecm, Nov 01 '20 at 21:24
If you can't do it in one instruction, then use more than one instruction. — Erik Eidt, Nov 01 '20 at 21:29
@ecm the one you provided is correct, I made a mistake I used as a example bwsap with a 32bit register eax — kannzzmm2, Nov 01 '20 at 21:46
[edit] your question to fix the expected input and output. Anyway, `bswap` works with 32 and 64 bit too. Does neither of those produce the result you want? — Jester, Nov 01 '20 at 22:52

Peter Cordes · Answer 1 · 2020-11-01T23:32:37.627

bswap eax will reverse the low 4 bytes of RAX, zeroing the top 4 bytes of RAX as part of writing the result to EAX. 32-bit operand-size is always available, you don't have to use 64-bit operand-size in 64-bit mode. Look at any compiler output for code using int and you'll see 32-bit registers being used. See also The advantages of using 32bit registers/instructions in x86-64

In fact bswap r32 is faster (1 uop, 1c latency) than bswap r64 (2 uops, 2c latency) on mainstream Intel (including even Ice Lake) https://uops.info/

The low 4 bytes forward and reverse are
00 66 6E 69 input
69 6E 66 00 32-bit bswap result

That gives the result you say you want, for this input. If you have some other requirement for larger inputs with non-zero high halves, say so. (If you have SSSE3, you can do an arbitrary byte shuffle on an XMM register; worth considering if a bswap r32 or bswap r64 isn't what you want.

How to swap bytes in 64bit Registers using NASM?

1 Answers1