On Thu, Oct 8, 2009 at 7:42 AM, Rick Brown <rick.brown.3@xxxxxxxxx> wrote:
> Hello list,
>
> As far as I recall from K&R, isn't pointer arithmetic on a void
> pointer banned? And any effort to do that results in an error -
> because the compiler won't know by how much size to increment the
> pointer for a statement like "ptr++"? But then how about this:
But in the program, you aren't actually trying to dereference the
value. Just adding means it becomes normal arithmetic and that is why
you get result as 1. You will see the error if you try to dereference
it.
/tmp> gcc a.c
a.c: In function ‘main’:
a.c:5: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘void *’
a.c:6: warning: dereferencing ‘void *’ pointer
a.c:6: error: invalid use of void expression
/tmp> cat a.c
#include <stdio.h>
int main()
{
void *ptr = 0;
printf("%d \n", ptr+1);
printf("%d \n", *(ptr+1));
}
>
> [rick@linux rick]$ cat t.c
> #include <stdio.h>
> int main()
> {
> void *ptr = 0;
> printf("%d \n", ptr+1);
> }
> [rick@linux rick]$ gcc t.c
> [rick@linux rick]$ ./a.out
> 1
> [rick@linux rick]$
>
> It compiles and runs fine ... !
>
> TIA,
>
> Rick
>
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>
--
Thanks -
Manish
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