linux-kernel - Re: [tip:x86/mce] x86/bitops: Move BIT

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Date:	Wed, 23 May 2012 19:11:25 +0200
From:	Borislav Petkov <bp@...en8.de>
To:	Linus Torvalds <torvalds@...ux-foundation.org>
Cc:	"H. Peter Anvin" <hpa@...or.com>,
	Peter Zijlstra <a.p.zijlstra@...llo.nl>,
	Borislav Petkov <borislav.petkov@....com>, mingo@...nel.org,
	linux-kernel@...r.kernel.org, frank.arnold@....com,
	akpm@...ux-foundation.org, tglx@...utronix.de,
	linux-tip-commits@...r.kernel.org
Subject: Re: [tip:x86/mce] x86/bitops: Move BIT_64() for a wider use

On Wed, May 23, 2012 at 09:57:47AM -0700, Linus Torvalds wrote:
> On Wed, May 23, 2012 at 9:53 AM, Borislav Petkov <bp@...en8.de> wrote:
> >
> > How about the following completely untested chunk:
> 
> No, you can't do that. All standard C operations will return *one*
> type. That very much includes the ternary ?: operator.

And, in addition, hpa's example won't work too:

	u64 msr = ~BIT(1);

> The *only* ways I know of to get two types are
> 
>  - C preprocessor stuff, ie
> 
>      #define BIT(x)  __BIT_##x
> 
>    and then just enumerate all the 64 cases. This is portable, but it gets old.

Nah, that's ugly.

>  - using __builtin_choose_expr(), which actually allows the two
> expressions to have different types, but requires a very strict
> compile-time constant (ie you cannot rely on the optimizer making it a
> constant - because it needs to choose the expression before the
> optimizer runs)
> 
> There might be some other magic gcc extension, of course.

Hmm and if not, it looks like BIT_64 is the easiest and most readable
thing we can do.

Oh well.

-- 
Regards/Gruss,
    Boris.
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