linux-kernel - Re: somebody dropped a (warning) bomb

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Date:	Thu, 8 Feb 2007 16:42:40 -0800 (PST)
From:	Linus Torvalds <torvalds@...ux-foundation.org>
To:	David Rientjes <rientjes@...gle.com>
cc:	Jan Engelhardt <jengelh@...ux01.gwdg.de>,
	Jeff Garzik <jeff@...zik.org>,
	Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
	Andrew Morton <akpm@...ux-foundation.org>
Subject: Re: somebody dropped a (warning) bomb

On Thu, 8 Feb 2007, David Rientjes wrote:
> 
> Your struct:
> 
> 	struct dummy {
> 		int flag:1;
> 	} a_variable;
> 
> should expect a_varible.flag to be signed, that's what the int says.  

No it's not.

You just don't understand the C language. 

And if you don't understand the C language, you can't say "that's what the 
int says". It says no such thing.

The C language clearly says that bitfields have implementation-defined 
types. So when you see

	struct dummy {
		int flag:1;
	} a_variable;

if you don't read that as "oh, the sign of 'flag' is implementation- 
defined", then you simply aren't reading it right.

Is it "intuitive"? Nobody ever really called C an _intuitive_ language. C 
has a lot of rules that you simply have to know. The bitfield sign rule is 
just one such rule.

> There is no special case here with regard to type.

Sure there is. Read the spec.

I don't understand why you are arguing. YOU ARE WRONG.

Bitfields simply have implementation-defined signedness. As do enums. As 
does "char". It really is that simple. 

The *real* special case is actually "int" and "long". In many ways, the 
fact that those *do* have a well-specified signedness is actually the 
exception rather than the rule.

Most C types don't, and some you can't even tell (do pointers generate 
"signed" or "unsigned" comparisons? I'll argue that a compiler that 
generates signed comparisons for them is broken, but it tends to be 
something you can only see with a standards- conforming proghram if you 
can allocate memory across the sign boundary, which may or may not be 
true..)

			Linus
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