Re: The type of bitops

[Linus Torvalds]

On Mon, May 6, 2013 at 6:36 PM, H. Peter Anvin <hpa@xxxxxxxxx> wrote:
>>
>> That sounds insane. Just *how* could the size of the argument even
>> matter? Seriously, there's no *difference* between a 32-bit or a
>> 64-bit index. The code is the same, there's no possible reason to make
>> it be different.
>>
>
> The only reason on x86 would be to avoid the REX prefix.  Hardly worth
> it, though.

Hmm. You *might* be able to do it by simply turning the inline
functions into statement expressions, and letting the compiler do the
types as it will. Use "(nr)+0" to make sure that the type is at least
"int", and then "Ir" as the asm constraint.

A quick test seems to imply that this works:

  #define test_bit(nr, addr) \
    ({ int oldbit; \
        asm volatile("bt %2,%1\n\t" \
                "sbb %0,%0" \
                : "=r" (oldbit) \
                : "m" (*(unsigned long *)(addr)), \
                 "Ir" ((nr)+0)); \
    oldbit; })

  int test(int idx1, long idx2, void *addr)
  {
    return test_bit(idx1, addr) + test_bit(idx2, addr);
  }

and I get

    ...
    bt %edi,(%rdx)
    ...
    bt %rsi,(%rdx)
    ...

for the int/long index respectively. So if the argument is an 'int',
it will avoid the rex prefix.

And I tested that using a "char" index does indeed do the right thing too:

    movsbl %dil,%eax
    bt     %eax,(%rcx)

so that all seems surmountable.Whether it really matters, I dunno.

                 Linus
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