From: Robert Love <rml@novell.com>
To: ncunningham@linuxmail.org
Cc: Linux Memory Management <linux-mm@kvack.org>
Subject: Re: Is sizeof(void *) ever != sizeof(unsigned long)?
Date: Sat, 04 Dec 2004 11:26:17 -0500 [thread overview]
Message-ID: <1102177577.6052.39.camel@localhost> (raw)
In-Reply-To: <1102155752.1018.7.camel@desktop.cunninghams>
On Sat, 2004-12-04 at 21:40 +1100, Nigel Cunningham wrote:
> I guess the subject line says it all.
In general? Sure. There is no guarantee in C or the "ABI Writer's
Handbook."
In Linux, though, especially the kernel, we run with that assumption.
An "unsigned long" can always hold a pointer, it is always equal to the
wordsize. This just means that architecture ports in Linux have to be
LP32 or LP64 or whatever.
A lot of code in the kernel uses an "unsigned long" instead of a "void*"
to hold a generic memory address. I personally like this practice, if
you never intend to directly dereference the pointer.
Robert Love
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next prev parent reply other threads:[~2004-12-04 16:26 UTC|newest]
Thread overview: 5+ messages / expand[flat|nested] mbox.gz Atom feed top
2004-12-04 10:40 Is sizeof(void *) ever != sizeof(unsigned long)? Nigel Cunningham
2004-12-04 16:26 ` Robert Love [this message]
2004-12-04 21:14 ` Nigel Cunningham
2004-12-04 17:02 ` Fawad Lateef
2004-12-04 21:32 ` Robert Love
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