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

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Date:	Wed, 23 May 2012 09:57:47 -0700
From:	Linus Torvalds <torvalds@...ux-foundation.org>
To:	Borislav Petkov <bp@...en8.de>, "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, torvalds@...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 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.

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.

 - 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.

                 Linus
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