Re: Using intptr_t and uintptr_t in Kernel

[Mikael Pettersson]

Leonidas . writes:
 > >  > 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

That's based on what C99 states, but it's not the definition of uintptr_t
but a specification of the minimal size of that type.  The page you're
referring to fails to mention that.

Anyway, this is not an issue with the kernel.  Please direct C questions
to a C-specific forum, like comp.std.c.  Or google n1124.pdf and read that.
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