GCC Inline Assembler- How to "BSWAP" the lower 32-bit of 64-bit register???

Question

Hello. I've been looking for the answer 'how to use 'BSWAP' for lower 32-bit sub-register of 64-bit register.' For example, "0x0123456789abcdef" is inside RAX register, and I want to change it to "0x01234567efcdab89" with a single instruction. (because of performance) So I tried following inline function:

#define BSWAP(T) {  \
    __asm__ __volatile__ ( \
      "bswap %k0" \
      : "=q" (T) \
      : "q" (T)); \
}

And the result was "0x00000000efcdab89". TT-TT I, actually, don't understand why the compiler acts like this. Does anybody know the efficient solution???

- A Korean boy, who haven't slept for 2 days to solve this problem. =_=

Answer 1

Check the assembly output generated by gcc! Use the gcc -s flag to compile the code and generate asm output.

IIRC, x86-64 uses 32-bit integers by default when not explicitly directed to do otherwise, so this may be (part of) the problem.

Answer 2

Thanks for your answer.

Um.. In the asm code, it was using 32-bit register, namely %ebx, as I intended. (you know, 'k' suffix in the GCC inline assembly code means it's 32-bit register.) And I tested some experiments with the asm code, but couldn't find the answer which is how to change the lower 32-bit sub-register without changing the upper 32-bit sub-register of 64-bit register. TT-TT So I thought that changing a specific part of register is not possible. However, the interesting one is that it works with lower 16-bit sub-register. Um.. I think it's better to say with examples. (Sorry, my English isn't good enough to talk clearly...-.-;;)

One of my tests was done by using ror instruction.

Suppose that a 64-bit variable has the value of "0x0123456789abcdef". (and it's in the rbx register.)

When the asm code was "ror $8, %rbx", the result was, of course, "0xef012345678abcd".

When the asm code was "ror $8, %ebx", The expected value was "0x01234567ef89abcd", but the result was, (TToTT), "0x00000000ef89abcd". The upper 32-bit value was gone. So I thought I can't change the only part of register.

However, when the asm code was "ror $8, %bx", the result was, (@o@), "0x0123456789abefcd", instead of "0x000000000000efcd". That's why I'm so confused.

Answer 3

Ah, yes, I understand the problem now:

the x86-64 processors implicitly zero-extend the 32-bit registers to 64-bit when doing 32-bit operations (on %eax, %ebx, etc). This is to maintain compatibility with legacy code that expects 32-bit semantics for these registers, as I understand it.

So I'm afraid that there is no way to do ror on just the lower 32 bits of a 64-bit register. You'll have to do use a series of several instructions...

Answer 4

Thanks a lot. Now, everything is clear. (^___^)