Le sam 19/07/2003 � 13:01, Shachar Shemesh a �crit :
> Todd Vierling wrote:
>
> >On Sat, 19 Jul 2003, Vincent B�ron wrote:
> >
> >: That's an easy one. One an Alpha, sizeof(int) != sizeof(foo *). I think
> >: sizeof(int) == 8, while sizeof(foo *) == 4.
> >
> >Other way round. sizeof(int) == 4; sizeof(void *) == 8.
> >
> >: gcc is kind enough to warn you about this potentially nasty situation,
> >: although in these particular cases I don't see how putting the pointer
> >: in something larger will create a problem.
> >
> >Use intptr_t. If that's not available (via autoconf test), typedef intptr_t
> >to be "unsigned long" locally, as "long" is the size of a pointer on all
> >typical ILP32 and LP64 hosts.
> >
> >
> >
> The reason something is a warning and not an error is that it MAY have a
> legitimate use. As such, there should always be a way to change the code
> so that it will keep the same meaning, but will avoid the warning in the
> future. In this particular case, for example, what does applying the
> following change do?
> - if (!((int)id >> 16)) return wine_dbg_sprintf( "<guid-0x%04x>",
> (int)id & 0xffff );
> + if (!((int)id >> 16)) return wine_dbg_sprintf( "<guid-0x%04x>",
> (UINT16)id );
You'll have to do it also at the beginning of the line.
>
> I'm a bit confused. gcc seems to be warning us about an explicit cast? I
> always thought that, in C, explicit casts are THE way of avoiding
> compile time warnings. Why should one change the semantic meaning of the
> code just so that gcc is pacified?
I think for pointers (especially when you change the size of the result)
gcc is more picky. Compile the attached program to verify.
And I don't remember Steven saying that he actually used gcc or even
Linux on the Alpha, it could be the DEC cc under Tru-64...
Vincent
#include <stdio.h>
#include <stdlib.h>
int toto(int *pointer) {
if ((short int)pointer >> 16)
return 0;
else
return 1;
}
int main(void) {
int b;
printf("%d\n", toto(&b));
exit(0);
return 0;
}
