Re: Using intptr_t and uintptr_t in Kernel

["Leonidas ." <leonidas137@gmail.com>]

On Tue, Oct 13, 2009 at 2:19 PM, Mikael Pettersson <mikpe@it.uu.se> wrote:
> 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.
>

Thanks, for the response, but frankly I am still confused.
Let me recollect my thoughts in more concise manner.


User space documentation for C99, http://linux.die.net/man/3/intptr_t says,

typedef unsigned int uint16_t
typedef uint16_t uintptr_t

and http://lxr.linux.no/#linux+v2.6.31/include/linux/types.h#L41 says,

typedef unsigned short __u16;
typedef         __u16           uint16_t;

i.e. kernel space mean 16 bits when it says u16 but gcc does not mean
the same thing.
So if a programmer meant uintptr_t in userspace he means store the
pointer in a datatype
which is as big as uint16_t i.e. unsigned int. So hold the pointer in
an unsigned int.

In kernel space if we go by the data type definitions, we would get a
pointer of size __u16
which is an unsigned short, the programmer did not intend this, right?


-Leo.