Re: [Linux-ia64] ia-64 cast warning??

[Dan Pop <Dan.Pop@cern.ch>]

On Wed, 14 Nov 2001, Pereira, D LailaX E wrote:

> Consider the following macro : executing in a program , on linux ia-64
> machine: 
> ifdef CIRC_DEBUG
> #define CBUFSIZE 32768 
> #define CHECK_CIRC_ALIGNMENT( A ) \
> if (((unsigned int)(circbuff + A))%sizeof(caddr_t)) { A += (sizeof(caddr_t)
> - (((unsigned int)(circbuff + A))%sizeof(caddr_t)))/sizeof(unsigned short);
> }
> 
> #define CIRCADDR( A ) \
> CHECK_CIRC_ALIGNMENT(CBptr) \
> *(caddr_t *)(circbuff + CBptr) = A; \
> CBptr = (CBptr + sizeof(caddr_t)/sizeof(short)) % CBUFSIZE;
> 
> #endif
> 
> 
> int main(){
>   unsigned int volatile* mask;
>   unsigned short circbuff[CBUFSIZE];
>   unsigned int CBptr;
>   unsigned int current;
> .....
>   CHECK_CIRC_ALIGNMENT(current);
> .....
> }
> 
> I always get the warning : 
> warning : cast from pointer to integer of different size  on the line
> CHECK_CIRC_ALIGNMENT(current);
> 
> when I compile the above program.I want to get rid of the warning.I am not
> able to figure out why I get this warning.??

The reason of the warning is OBVIOUS: you are converting a pointer
value (circbuff + current), which is a 64-bit value, to unsigned int,
which is a 32-bit type.  The compiler is warning you that you may lose
significant bits in this conversion, and this is the case, indeed, on
Linux-ia64.

As you have already been told, if you need to make such conversions,
use the type uintptr_t, defined in <stdint.h>, instead of some
arbitrary, hard coded integer type.

Dan