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From: ark@alice.UUCP (Andrew Koenig)
Newsgroups: comp.lang.c
Subject: Re: Ambiguous C?
Message-ID: <9249@alice.UUCP>
Date: 27 Apr 89 14:25:30 GMT
References: <111@ssp1.idca.tds.philips.nl>
Organization: AT&T Bell Laboratories, Liberty Corner NJ
Lines: 40

In article <111@ssp1.idca.tds.philips.nl>, roelof@idca.tds.PHILIPS.nl (R. Vuurboom) writes:

> The following piece of code got me into trouble.
> I needed to access a register which only allows long accesses.

 . . .

> Does C specify which (if any) interpretation is correct?

> #define	SCUCMD		0 
> #define	SCU_BDID	0x2000		/* bit 14 */

> 	if ( (*(int *)(SCUCMD)) & SCU_BDID )

There are two separate questions:

	1.  Does C define the meaning of this program fragement?

	2.  How do you get the compiler to do what you want?

The answer to (1) is `no' -- once you cast an integer into
a pointer, you're on your own.

In practice, what's probably happening is that the compiler
realizes that there's only one bit turned on in the value
of SCU_BDID, so it's optimizing the code by not fetching
the entire word to test.

To suppress the optimization, you can probably conceal the
value of SCU_BDID from the compiler by using it in a way that
the compiler thinks might change.  For instance:

	static int SCU_BDID = 0x2000;

Now the compiler probably won't be smart enough to realize that
SCU_BDID is really a constant, so it will have to generate code
that caters to any bit in the whole word being on.
-- 
				--Andrew Koenig
				  ark@europa.att.com