On Wed, Jan 06, 2010 at 07:29:21PM +0000, Joshua Haberman wrote:
> Erik Trulsson <ertr1013 <at> student.uu.se> writes:
> > > int i;
> > > unsigned int *pui = (unsigned int*)&i;
> > > unsigned int ui = *pui;
> >
> > (First I will assume that 'i' will be assigned some value, to make sure it
> > does not contain a trap-representation, or the assignment to 'ui' would have
> > undefined behaviour.)
> >
> > I think 6.2.5 clause 27 is very relevant for this. It says that 'pointer to
> > int' and 'pointer to union' do not need to have the same representation as
> > each other. It also seems that 'pointer to int' and 'pointer to unsigned
> > int' do not need to have the same representation requirements (at least I
> > cannot find anything that says that signed and unsigned variants are
> > compatible types.) (Which I must admit comes as a bit of a surprise to me.)
> >
> > So, yes, that example does technically seem to be undefined (but I don't
> > know of any real-world implementation where it would not work as expected.)
>
> I am wondering how, under this interpretation, an "int" and "unsigned
> int" could ever alias each other, as 6.5p7 says they can. I believe now
> (especially after the notes Nick pointed us to) that your interpretation
> of 6.5p7 is the intended one. But if that is the case, then we should
> expect to find some legal way in which "int" and "unsigned int" could
> come to be aliased. With your interpretation about pointer conversions,
> I don't see how this could be.
Would not the following be a legal way of having that happen:
#include <stdlib.h>
void foo(int *i, unsigned int *ui)
{
...
}
int main(void)
{
void *pv;
int *pi;
unsigned int *pui;
pv=malloc(sizeof(int)); /* Assume that the malloc succeeds */
pi = pv;
pui = pv;
foo(pi, pui);
}
I believe that is legal code where '*i' and '*ui' would refer to the same
object (i.e. be aliased). (See 6.5p6 and 6.5p7)
--
<Insert your favourite quote here.>
Erik Trulsson
ertr1...@student.uu.se
