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Message-ID: <19156.16146.326389.380625@pilspetsen.it.uu.se>
Date: Tue, 13 Oct 2009 10:49:22 +0200
From: Mikael Pettersson <mikpe@it.uu.se>
To: "Leonidas ." <leonidas137@gmail.com>
Cc: linux-kernel <linux-kernel@vger.kernel.org>
Subject: Re: Using intptr_t and uintptr_t in Kernel
In-Reply-To: <f22d86810910130037x48cac5c1s8928a6ff49dda381@mail.gmail.com>
References: <f22d86810910130037x48cac5c1s8928a6ff49dda381@mail.gmail.com>
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Leonidas . writes:
 > I know variants of this questions have been asked on this list before,
 > but I could not
 > find a thread where a conclusion was reached, most of the time
 > responses were { I don't care,
 > developer's choice, might be in future } and variations thereof. But
 > one thing was clear
 > that do not mix userspace and kernel space data types.
 > 
 > Currently, I am porting a user space application to kernel, this app
 > uses intptr_t data type.
 > 
 > A look at linux/types.h shows that,
 > 
 > typedef unsigned long           uintptr_t;
 > 
 > but intptr_t is not defined at all. Also, isn't above definition
 > incorrect?

No, it's correct because Linux requires sizeof(void*) == sizeof(long).

 > Since the whole idea
 > behind uintptr_t is to store pointer in a int sized variable,

uintptr_t will use _some_ integer type, not necessarily 'int'.

If ISO C said 'int' there would be no need for {,u}intptr_t.

 > are we
 > not assuming here that
 > 
 > sizeof(int) = sizeof(unsigned long ) on all archs?

No, see above.
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